Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Chapter 1.1 These problems implicitly make use of the following lemma (stated casually in the book).
a b a b a b a b a b
Lemma If Kß æ is a group and L © K is closed under
æ
, then
æ
is associative in L .
Proof. We know know æ is assoc associat iative ive in K by defin definiti ition on of a grou group. p. Now let let +ß ,ß - − L. Since Since L is closed, + æ , æ - ß + æ , æ - − L . However, since +ß ,ß - − L implies +ß ,ß - − K we also know + æ , æ - œ + æ , æ - . Hence, æ is associative in L .
Problem 1.1.1 Determine which of the following binary operations are associative:
(a) the operation æ on ™ defined +æ , œ + , (b) the operation æ on ‘ defined by +æ , œ + , +, +, &
(c) the operation æ on defind by +æ , œ
a b a b a ef a b a b a b a b a a b a b a b a b a b aa b a bb a b a b a b aa b a a b a b a bb a a b aa b a bb a b a b a b a b
(d) the operation æ on ™ ‚ ™ defined by +ß ,
- ß . œ +. , - ß ,. (e) the operation æ on ! defined by a æ b œ +, . æ
Solution. (a) Not associative. For example, #æ "
æ"
æ-
œ + , +,
æ-
a b
œ ! Á # œ #æ "æ " .
(b) Associative, because
+æ ,
b
b
œ + , +, - + , +, -
a b a b
œ + , - , - + , - , - œ + , æ - + ,æ - œ + æ , æ - .
The intermediate steps follow because usual addition and multiplication multiplicat ion is associative and commutative in ™. (c) Not associative. For example, !æ !
æ #&
œ & Á " œ !æ !æ #& .
(d) Associative, because
+ß ,
æ
-ß .
æ
/ß 0 œ +. + . ,- ß ,. ,.
/ß 0 œ
æ
œ + . 0 , - 0 . / ß , .0 .0 œ +ß ,
æ
-ß .
æ
a b a bb ba b
+. ,- 0 ,. ,. /ß ,. ,. 0 œ +ß ,
æ
- 0 . /ß . 0
/ß 0 .
Notice we could not say ™ ‚ ™ß æ is isomorphic to ß even though intuitively because this would exclude , or . equal to ! (which is encompassed by the former). (e) Not associative. For example, "æ "
æ#
œ
" #
+ ,
.
œ
+, , ,. ,
Á # œ "æ "æ # .
Problem 1.1.2 Decide which of the binary operations in the preceding exercise are commutative. Solution. (a) Not commu commutat tative ive.. For For exampl example, e, 1 æ ! œ " Á " œ !æ ".
(b) Commutative, Commutativ e, because
+æ , œ + , +, œ , + ,+ œ ,æ + , due to addition and and multiplication multiplicat ion being commutative in ™. (c) Commutative, Commutati ve, because
+æ , œ
+, &
œ
, + &
œ ,æ +,
due to addition being commutative in ™.
Robert Krzyzanowski
Solutions to Dummit and Foote
(d) Commutative, Commutativ e, because
a ba b a +ß ,
æ
b a
Chapter 1
b a ba b
- ß . œ +. ,- ß ,. , . œ - , . +ß ., ., œ -ß .
æ
+ß , ,
due to addition and and multiplicaiton multiplicai ton being commutative in ™. (e) (e) Not Not comm commut utat ativ ive. e. For For exam exampl ple, e, #æ " œ # Á
" #
œ "æ #.
Problem 1.1.3 Prove that the addition of residue classes in ™Î8™ is associative (you may assume it is well defined). Proof .
ˆ ‰ ˆ ‰
,ß - − ™Î8™. Then + , - œ + , - (by definition--see page 9 in the book), Let +ß ,ß
a b a b
which equals + , - œ + , - (again by definition). However,
+ , - œ + , - œ + , - œ + , -. Problem 1.1.4 Prove that the multiplication of residue classes in ™Î8™ is associative (you may assume it is well defined).
ab
Proof .
ˆ ‰
,ß - − ™Î8™. Then + , † - œ + † ,- (by definition--see page 9 in the book), which Let +ß ,ß
equals + ,- œ +,- (again by definition). However,
ˆ ‰
ab
+,+,- œ +, - œ +, † - œ + † , - . Problem 1.1.5 Prove for all 8 " that ™Î8™ is not a group under multiplication of residue classes.
a b
‚
Proof . In the book, we've seen ™Î8™ is a group. Hence, ! must be the guilty element of breaking this structure. Indeed, ! has no inverse, since ! † + œ ! † + œ ! œ + † ! œ + † ! for any + − ™Î8 Î 8 ™, and we know " is the identity since " † + œ " † + œ + œ + † " œ + † " for any + − ™Î8 ™. Since there is no element + such that ! † + œ ", ! has no inverse and by definition ™Î8™ is not a group under multiplication multiplicati on of residue classes.
Problem 1.1.6 Determine which of the following sets are groups under addition:
(a) the set of rational numbers (including ! œ !Î") in lowest terms whose denominators are odd (b) the set of rational numbers (inculding ! œ !Î") in lowest terms whose denominators are even (c) the set of rational numbers of absolute value " (d) the set of rational numbers of absolute value " together with ! (e) the set of rational numbers with denominators equal to " or # (f) the set of rational numbers with denominators equal to ", #, or $. Solution. For each respective problem, call the group K.
(a) This is a group. First, if
+ ,ß .
a b a b
− K with # Îl , and # Îl . (i.e., both are odd) and +ß , œ - ß . œ ", then + ,
.
œ
+. ,,.
−K
since # Îl ,.; hence we have closure. We know it is associative since is on addition and K § . The identity is !Î" since !" +, œ +, œ +, !" for all +, − K. Finally, each element has an inverse since + + ! + , , œ " for any , − K. (b) This is not a group because it does not have closure. For example, lowest terms and the denominator is odd (not even).
" #
− K but "# "# œ
" "
 Ksince "" is in
Robert Krzyzanowski
Solutions to Dummit and Foote
(d) Commutative, Commutativ e, because
a ba b a +ß ,
æ
b a
Chapter 1
b a ba b
- ß . œ +. ,- ß ,. , . œ - , . +ß ., ., œ -ß .
æ
+ß , ,
due to addition and and multiplicaiton multiplicai ton being commutative in ™. (e) (e) Not Not comm commut utat ativ ive. e. For For exam exampl ple, e, #æ " œ # Á
" #
œ "æ #.
Problem 1.1.3 Prove that the addition of residue classes in ™Î8™ is associative (you may assume it is well defined). Proof .
ˆ ‰ ˆ ‰
,ß - − ™Î8™. Then + , - œ + , - (by definition--see page 9 in the book), Let +ß ,ß
a b a b
which equals + , - œ + , - (again by definition). However,
+ , - œ + , - œ + , - œ + , -. Problem 1.1.4 Prove that the multiplication of residue classes in ™Î8™ is associative (you may assume it is well defined).
ab
Proof .
ˆ ‰
,ß - − ™Î8™. Then + , † - œ + † ,- (by definition--see page 9 in the book), which Let +ß ,ß
equals + ,- œ +,- (again by definition). However,
ˆ ‰
ab
+,+,- œ +, - œ +, † - œ + † , - . Problem 1.1.5 Prove for all 8 " that ™Î8™ is not a group under multiplication of residue classes.
a b
‚
Proof . In the book, we've seen ™Î8™ is a group. Hence, ! must be the guilty element of breaking this structure. Indeed, ! has no inverse, since ! † + œ ! † + œ ! œ + † ! œ + † ! for any + − ™Î8 Î 8 ™, and we know " is the identity since " † + œ " † + œ + œ + † " œ + † " for any + − ™Î8 ™. Since there is no element + such that ! † + œ ", ! has no inverse and by definition ™Î8™ is not a group under multiplication multiplicati on of residue classes.
Problem 1.1.6 Determine which of the following sets are groups under addition:
(a) the set of rational numbers (including ! œ !Î") in lowest terms whose denominators are odd (b) the set of rational numbers (inculding ! œ !Î") in lowest terms whose denominators are even (c) the set of rational numbers of absolute value " (d) the set of rational numbers of absolute value " together with ! (e) the set of rational numbers with denominators equal to " or # (f) the set of rational numbers with denominators equal to ", #, or $. Solution. For each respective problem, call the group K.
(a) This is a group. First, if
+ ,ß .
a b a b
− K with # Îl , and # Îl . (i.e., both are odd) and +ß , œ - ß . œ ", then + ,
.
œ
+. ,,.
−K
since # Îl ,.; hence we have closure. We know it is associative since is on addition and K § . The identity is !Î" since !" +, œ +, œ +, !" for all +, − K. Finally, each element has an inverse since + + ! + , , œ " for any , − K. (b) This is not a group because it does not have closure. For example, lowest terms and the denominator is odd (not even).
" #
− K but "# "# œ
" "
 Ksince "" is in
Robert Krzyzanowski
Solutions to Dummit and Foote
kk
Chapter 1
(c) This is not a group because it does not have closure. For example, " " Î ". # # œ "  K since "
¸¸
(d) This is not a group since it fails closure. For example, " " Î " and "# Á !. # Â K since #
$ # ß "
− K since
" #
¸¸ " #
− K since
¸¸ k k $ #
" ", but
$ #
Ÿ " but
a b
" œ
(e) Assume each rational number is in lowest form. This is a group. First, take +, ß .- − K with the greatest common divisor of + and ,, and - and . equal to 1. Consider , œ . œ #. Then +# #- œ + # − K. Otherwise, + +. , , . œ , . − K since , . œ " or # since gcd , . ß +. ,- œ " or # (if , . œ # and +. ,- is even, the gcd becomes "). Hence, K is closed. We know it is associative since is on addition and K § . The identity is !Î" since !" +, œ +, œ +, !" for all +, − K. Finally, each element has an inverse since + + ! + , , œ " for any , − K. By definition, K is a group.
a
b
(f) This is not a group because it is not closed. For example,
e c d cd c d
f
" " #ß $
− K but
" #
" $
œ
" '
 K.
Problem 1.1.7 Let K œ B − ‘ l ! Ÿ B " and for Bß C − K let B æ C be the fractional part 90 B C (i.e., Bæ C œ B C B C where + is the greatest integer less than or equal to +). Prove that æ is a welldefined binary operation on K and that K is an abelian group under æ ( called the real numbers mod ").
c d
c d
Proof . To show the operation is well-defined, notice either ! Ÿ B C 1 or " Ÿ B C #. In the former case, B C œ ! so that Bæ C œ B C B C œ B C . Otherwise, B C œ " so that Bæ C œ B C ". Hence, æ is a well-defined binary operation on K. To show closure, again consider the two cases mentioned earlier. In the former, Bæ C œ B C and since ! Ÿ B C œ Bæ C 1 by assumption, Bæ C − K. In the latter case, Bæ C œ B C " and since " Ÿ B C # we have " " œ ! Ÿ B C " œ Bæ C # " œ " so again Bæ C − K. Therefore, K is closed. To show it is associative, notice for Bß Cß D − K,
a b a Bæ C
æD
c db a c d c c d c c d c d c
œ BC BC
æD
c db c c d d c d a b c a bd a b c d d c d c c d d c d
œ BC BC D BC BC D œ
œ B C D C D B C D C D œ B C æ D B C æ D œ Bæ C æ D . The middle equality holds because B C B C B C D œ C D B C D C D which needs to be explicitly justified case-by-case. Assume ! Ÿ B C Ÿ " and ! Ÿ C D Ÿ ", or " Ÿ B C # and " Ÿ C D #. Then B C œ C D œ " so the equation holds. Otherwise, assume without loss of generality ! Ÿ B C Ÿ " and " Ÿ C D #. Then B C œ ! and C D œ ", so that
c d c
c d d c
d
c
d c d c
c d
B C B C B C D œ B C D œ " B C D " œ C D B C D C D .
c d c d c d a b c a bd a b ca b d c d c d e f
Hence, the operation is associative. Furthermore, ! is the identity since ! B ! B œ B ! B ! œ B B for any B − K . Finally, each element has an inverse, since B B B B œ B B B B B œ ! ! œ ! for each B − K . Therefore, K is a group. Finally, K is abelian since for any Bß C − K, we have that B C B C œ C B C B since addition is commutative in ‘. Hence, K is an abelian group.
c d
Problem 1.1.8 Let K œ D − ‚ l D 8 œ " for some 8 − ™ .
(a) Prove that K is a group under multiplication (called the group of roots of unity of ‚). (b) Prove that K is not a group under addition.
a b a ba b
To prove closure, let Aß D − K. Then b8ß 7 − ™ such that A8 œ D 7 œ ". Then AD 87 œ A8 7 D 7 8 œ "7" 8 œ " and since 87 − ™ (the positive integers are closed under multiplication), multiplicat ion), by definition AD − K. Hence, K is closed. Associativity Associativit y is guaranteed since ‚Ï ! is a group under multiplication, multipli cation, and K § ‚Ï ! (notice !  K since there is no 8 − ™ such that !8 œ "). The identity Proof .
(a)
ef
ef
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
is " since for 8 œ " − ™ we have "" œ " so that " − K, and furthermore for all D − K, " † D œ D † " œ D. Finally, each element has an inverse since for each D − K there is an 8 − ™ such that D 8 œ ", so that D 8" D œ D † D 8" œ D 8 œ ". Therefore, Kß † is a group.
a b
(b) Since " − K, it can not be a group under multiplication multiplicat ion since " " œ #  K as there is no 8 − ™ such that #8 œ " (and hence K is not closed). Problem 1.1.9 Let K œ
š È +,
›
# − ‘ l +ß +ß , − .
(a) Prove that K is a group under addition. (b) Prove that the nonzero elements of K are a group under multiplication.
Š È ‹ Š È ‹ a b a bÈ aÈb a b Š È‹ Š È‹ Š È‹ È Š È‹ È È Š È ‹ Š a b È ‹ Š a bÈ ‹ Š È ‹ È È È Š È ‹Š È ‹ a b a b È a bÈ a b Š È ‹Š È ‹ Š È ‹ È Š È‹ È È Š È ‹Š È‹ Š È ‹Š È ‹ È È ef ef e f e f Proof .
(a) Let + ,
È
#ß - .
È
# − K. Then + ,
# -.
# œ +- ,.
#
is in the group, because + - ß , . − (since ß is a group). Associativity Associativit y is guaranteed since K § ‘ and
‘ß
!!
+,
is a group. The identity is ! !
# œ+,
# for any + ,
# − K,
+,
Let + ,
#ß - .
‘ß †
"!
# − K. Then
+,
#
+,
is a group. The identity is " !
# œ+,
# − K, + ,
# for any + , #
+ +# #,#
# +,
# œ +,
#
# œ + ,
# +,
# œ !!
#.
group, because +- #,. ß +. ,- − (since ß † and
!!
# − K. Finally, each element has an inverse since for
# + ,
Therefore, K is a group under addition. (b)
# since
, + ##, #
-.
# œ ++ - #,. +. +. ,-
# is in the
is a group). Associativity Associativit y is guaranteed since K § ‘
# since
"!
#
+,
# œ +,
# †
# − K. Finally, each element has an inverse since for # œ
+ + ##, #
, + ##, #
#
+,
# œ "!
# (this was
obtained by solving for - and . in +- #,. œ "ß +. ,- œ !). We know +# +#, # +# ,#, # # − K since + , +# #, # ß +# #,# − K (notice the denominator can never be ! or that would contradict +ß , − , and neither can both the terms be ! since !  KÏ ! ). Therefore, KÏ ! is a group under multiplication. multiplicati on. Problem 1.1.10 Prove that a finite group is abelian if and only if its group table is a symmetric matrix.
Proof . Assume K œ 1" ß ÞÞÞß 1 8 is a finite abelian group. Then the 3, 4 entry in its group table is the group element 13 14 œ 14 13. The 4ß 3 entry in its group table is the group element 1 41 3 œ 1 31 4. Hence, by definition, the group table is a symmetric matrix. Now assume K œ 1" ß ÞÞ Þ ÞÞß 18 is a finite group with a symmetric matrix. Then the 3, 4 entry is the same as the 4ß 3 entry, that is, 1 31 4 œ 1 41 3. However, this holds for any two elements 13 ß 14 − K so that 1 3 14 œ 1 41 3 for all elements of K. This is precisely the definition of an abelian group.
Problem 1.1.11 Find the orders of each element of the additive group ™Î"#™ .
˜ ™ k k a k kb
Solution. The group is ™Î"#™ œ !ß "ß #ß ÞÞÞß "" . Then the orders, respectively, are ", "#, ', %, $, "#, #, "#, $ß %ß 'ß and "#. Notice these are K Îgcd Bß K . Indeed, this will be proven later.
a b
Problem 1.1.12 Find the orders of the following elements of the multiplicative group ™Î"#™ , (, "$.
‚
: " ß ", &, (
Robert Krzyzanowski
_
B
Solutions to Dummit and Foote
Chapter 1
ef
Solution. The identity is ", so the order for B is the smallest 8 − ™ _ such that B8 œ " (with œ "). Respectively, these are "ß #ß #ß #ß %, and " (since "$ œ ").
Problem 1.1.13 Find the orders of the following elements of the additive group ™Î$'™: "ß# , ' ,* ,"!ß "#ß "ß "!ß ").
ef
Solution. The identity is !, so the order for B is the smallest 8 − ™ _ such that 8B œ ! (with _B œ !). Respectively, these are $'ß ")ß 'ß %ß ")ß $ß $'ß ")ß #.
a b
Problem 1.1.14 Find the orders of the following elements of the multiplicative group ™Î$'™ "$ß "$ß "(. _
B
‚
: "ß "ß &ß
ef
Solution. The identity is ", so the order for B is the smallest 8 − ™ _ such that B8 œ " (with œ "). Respectively, these are "ß #ß 'ß $ß 'ß #.
a
Problem 1.1.15 Prove that +" +# ÞÞÞ+ 8
a b a
bb
b
"
" œ +8"+8" ÞÞÞ+ ""
aa
bb
a
"
b
Assume +" +# ÞÞÞ+ 8 B œ " so that B œ + "+ #ÞÞÞ+ 8 . Then +"" +" +# ÞÞÞ+ 8 B œ + "" † " so that +"" +" +# +$ÞÞÞ+ 8 B œ + #+ $ÞÞÞ+ 8 B œ + "" . Similarly, + $+ %ÞÞÞ+ 8 B œ + #"+ "". Applying this 8 times results " in B œ +8" +8" ÞÞÞ+ "" , as desired.
a ba
Proof.
kk
Problem 1.1.16 Let B be an element of K . Prove that B# œ " if and only if B is either " or #.
kk
kk
kk
Assume B# œ ". If B œ ", then B œ ". Otherwise, B Á " (only the identity has order ") so that B œ # by definition since # would be the smallest power B need be raised to in order to obtain the identity. On the other hand, assume B is either " or #. If B œ ", then B œ " as only the identity has order ". Otherwise B œ # so by definition of order, B# œ ". Proof .
kk
kk
kk
kk
Problem 1.1.17 Let B be an element of K . Prove that if B œ 8 for some positive integer 8 then B" œ B8" .
kk
Proof . Let B œ 8. By definition, B8 œ ". Hence, B † B8" œ B 8" † B œ " . This is precisely the definition of B" œ B8".
Problem 1.1.18 Let B and C be elements of K . Prove that BC œ CB if and only if C "BC œ B if and only if B" C "BC œ " . Proof . Assume Bß C − K and BC œ CB . Multiplying by C " on the left, C "BC œ C "CB œ B. Now assume C" BC œ B. Multiplying by B " on the left, B "C "BC œ B "B œ ". Finally, assume B "C "BC œ ". Multipling by CB on the left, CB B"C "BC œ C B † B " C "BC [generalized associativity] œ C † C " BC œ BC œ " † CB œ CB.
ab
ab
a b
Problem 1.1.19 Let B − K and let +ß , − ™ .
ab
(a) Prove that B+, œ B+ B, and B+
ab
(b) Prove that B+
"
œ B+ .
,
œ B+, .
(c) Establish part (a) for arbitrary integers + and , ( positive, negative, or zero).
a b
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Proof . Notice it is obvious B+ œ B+" B for all + − ™. This is because we can recursively define B +. If + œ !, then B+ œ ". Otherwise, B+ œ B+" † B.† Similarly, B + œ B +" † B " .
(a) We will induct on + and , using strong induction. First, notice B"" œ B# œ B † B [definition] œ B "B ". Now assume B87 œ B8B 7 for all 7 Ÿ 8 and 8 Ÿ 5 for some 5 − ™ . Then inductively we show B 5" 7 œ B5"B7 for all 7 Ÿ 5 ". First, B 5" œ B 5B " so that B 5"B œ B 5B B œ B 5 BB œ B 5B # [definition] œ B5#. The last step follows because if 5 œ ", B"B# œ BBB œ B $ œ B "#. Otherwise, we use our inductive assumption. Since B5# œ B 5" ", we have shown B 5" " œ B 5"B. Now assume B 5" ; œ B5" B; ; Ÿ 5. B 5"B ;" œ B 5"B ;B œ B 5" ;B for some Then [inductive 5" ;" 5" ;" assumption] œ B . Bœ B
ˆ ‰ ab
a b
a b
a b
a b
ab b a a bb
a ba b
a b
a ab b
a b a b a b a b a b a ba b a ba b ˆ ‰ a b ˆ‰ ˆ ‰ ˆ ‰ ab a b‰ ˆ ‰ ˆ a a bb a b ˆ ‰ a b a b k k kk k k akk b a b k k k k "
,
a b
7
Similarly, we can show B+ œ B+, . First, notice B" œ B"†". Now assume B 8 œ B 87 for all 7 Ÿ 8 and 7 " 8 Ÿ 5 for some 5 − ™ . Then inductively we show B8" œ B 8" 7 for all 7 Ÿ 8 ". First, B 8" œ
a b a b a b a
5
B 8" " . Now assume B8" œ B 8" 5 for some 5 Ÿ 8. Then B 8" B 8" 5 B 8" œ B 8" 5 8" [part (a)] œ B 8" 5" . (b)
As in part (a), we can show this inductively. First,
a b
a b
B 5" œ B 5" " † B " œ B 5 † B " œ B 5 " œ B+ in general. Hence, B+
ab
" "
B
œ B † B5
"
B"
"
5"
5
œ B 8"
œ B". Assume [Proposition
B 8" [part (a)] œ
B5
"
œ B5. Then " 1.1.1(4)] œ B 5" . +
(c) Let + be any integer. Then B+! œ B!+ œ B+ œ B!B + œ B +B !, and B !†+ œ B +†! œ " + œ B ! œ " ! œ B+ ! . Hence, part (a) is valid when + or , is zero. Otherwise, consider when + and , are negative. Then we
ab
know B
a bœB +,
"
+ ,
"
œ B,B + by part (a). Then B +, œ B ,B + and using part (b) and " " B, œ B + B , so that B +, œ B +B ,. Now Proposition 1.1.1(4), this yields B +, œ B+ without loss of generality assume + is positive and , is negative. Consider + , . Then + œ + , , with both parts positive. Hence, B+ B, œ B +, , B, œ B +, B,B , œ B +,. Now assume + , . Then + œ + , , with , , + , negative, and , + , , so by what we have just proved, B+ œ B+, B, . Therefore, B+ B, œ B +, B, B , œ B +, . B
a b
Problem 1.1.20 For B an element in K show that B and B" have the same order.
""
kk
a b
ab
8
"
Proof . Assume B œ 8 − ™ . By part (b) of the previous exercise, B" œ B 8 œ B 8 œ œ ". All that remains to be shown is that this is the least 8. Assume there is a 7 − ™ such that 7 8
a b kk a b ab 7
a b k k ˆa b ‰ k k ˆ ‰ a b a b k k " "
"
and B" œ ". Then B7 œ " so that B7 œ B 7 œ " " œ ". However, this would contradict the assumption B œ 7. Hence, B" œ 8. Now assume B œ _. Suppose B " has finite order, 8. Then 8
œ B8
"
"
œ " so again B8 " œ B 8 œ " " œ ". However, this would mean B has finite order, a contradiction. Therefore, B" œ _. B"
Problem 1.1.21 Let K be a finite group and let B be an element of order 8. Prove that if 8 is odd, then
ab
B œ B#
5
for some 5 .
If 8 œ ", B is the identity so the problem is trivial. Otherwise, let 7 − ™ be such that 7 7 #7" #7 8 œ #7 ". Then by the previous exercise, B" œ " so that B" œ B" œ B " B " œ ".
a b a b
Proof .
Multiplying by B on the right hand side, B"
†This
is justified because B+ œ
# Œ# +
+"
B B œ B+"B.
Bœ
3œ"
3œ"
#7 "
a b a b a b a b
B B œ B"
#7
œ B #7
"
œ B. Then
a b Ša b ‹ B #7
" "
œ
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
a b a b a b a b a b a b a b a ba b a b kk k k kk kk a b a b a b a b a ba b a b kk a b a b kk kk k k kk a b kk k k ka ba ba bk k k a b k k kk a b 7
7
"
B#7 œ B" . But by the previous exercise, B#7 œ B # . Hence, B " œ B # so that B" œBœ " 7 7 5 B# œ B# œ B# for 5 œ 7. If we wish to have a positive 5, let < be the least positive 5 5 5 <8 5<8 #< integer such that <8 7. Then B œ B# † "#< œ B# † B 8 . œ B# B # œ B #
ˆa b ‰ a b
Problem 1.1.22 If B and 1 are elements of the group K , prove that B œ 1" B1 . Deduce that +, œ ,+ for all +ß, − K .
First, we will show that for all 8 − ™, 1" B1
a b a b a b Proof 1.
1" B1
"
œ 1" B" 1. Otherwise, assume 1" B1 1" B1
5"
8
5
œ 1" B1
5
8
œ 1" B81. Inductively, when 8 œ " we have
œ 1" B5 1. Then
1" B1 œ 1" B5 1 1" B1 œ 1" B5 11" B1 œ 1"B5 B1 œ 1"B 5"1. 8
Hence, 1" B1 œ 1" B8 1 for 8 − ™. Then if B œ 8 − ™ we see 1 "B1 œ 1 "B 81 œ 1 "1 œ ". Now 7 assume there is a 7 − ™ such that 7 8 and 1" B1 œ ". Then 1" B71 œ " so left and right multiplication by 1 and 1 ", respectively, yields 11" B7 11" œ 11" so that B7 œ ". However, this contradicts the assumption B œ 8. Now assume that B œ _. If 1 " B1 œ 8 _ œ B , then
1" B1
k k kk
8
œ 1" B8 1 œ "
so again left and right multiplication by 1 and 1", respectively, yields B8 œ ", contradicting B œ _. Hence, 1" B1 œ B for all elements Bß 1 − K. To see the last part, let B œ +, and 1 œ , "+. Then
+, œ ,+" +, , " + œ ,+ .
kk kk k k kk
Proof 2.† As in Proof 1, we know 1" B1 when B œ _. If B œ 8, then
1" B1
8
8
œ 1" B8 1. First, we show 1"B1 Ÿ B . This is obvious
œ 1" B8 1 œ 1" 1 œ ",
so that 1" B1 Ÿ B for all Bß 1 − K. To see the opposite direction, substitute 1"B1 for B and 1 " for 1 in the previous statement:
¹a b a b ¹ k k k k 1"
kk k k
"
1" B1 1" œ B Ÿ 1"B1 .
ab
Thus, B œ 1" B1 for all Bß 1 − K. To see the last part of the problem, just notice ,+ œ + " +, + (e.g., take B œ +, and 1 œ +).
kk
kk ab
Problem 1.1.23 Suppose B − K and B œ 8 _ . If 8 œ => for some positive integers = and > , prove that B= œ >.
ab
>
kk ab a b a ba b a b ab
Proof . First, it is clear B= œ B=> œ B8 œ ". Assume there is a 7 − ™ such that 7 > and B= 7 œ ". Then B=7 œ ", but =7 => œ 8 , so this contradicts the fact B œ 8.
Problem 1.1.24 If + and , are commuting elements of K , prove that +,
ˆ ‰
ab
"
ab
5
8
œ + 8 , 8 for all 8 − ™. 5"
5
Inductively, +, œ +" , ". Assume +, œ +5 , 5 . Then +, œ +, +, œ + 5 , 5 +, œ +5 + ,5 , œ + 5", 5". Notice the penultimate step is justified by commutativity of + and ,. Hence, +, 8 œ +8 ,8 for all 8 − ™. When 8 œ !, +, ! œ " œ +!, !. Finally, when 8 − ™ , we know +, 8 œ + 8, 8 by Proof .
† This
ab
somewhat more elegant proof is due to Anssi Lahtinen from Stanford: math.stanford.edu/~lahtinen/math-120-f08/hwsolutions/math-120-sol-01.pdf
Robert Krzyzanowski
ab a
Solutions to Dummit and Foote
b a ba b a b ab
what we have just shown, but + and , commute so +, 8 œ ,8 +8 " œ +8 " , 8 " œ + 8, 8.
ab ab +,
8
œ ,+
8
Chapter 1
œ , 8 +8. Then
aa b b +,
8 "
œ
Problem 1.1.25 Prove that if B# œ " for all B − K then K is abelian. #
Proof 1. Let Bß C − K. Then BC œ " since BC − K . Hence, BCBC œ " so BC œ C "B ". However, # notice that CB œ " since CB − K, so CBCB œ " and consequently CB œ B "C ". Then BC CB œ C" B"B"C " œ ". This implies BC œ B "C " œ CB . Hence, BC œ CB for all Bß C − K and by definition K is abelian.
ab
a ba b
#
Proof 2. Let Bß C − K . Then BC œ BCBC œ " . Multiplying by B on the left and C on the right and noting that B# œ " and C # œ ", we obtain CB œ BC.
a b
Problem 1.1.26 Assume L is a nonempty subset of Kß æ which is closed under the binary operation on K and is closed under inverses, i.e., for all 2 and 5 − L , hk and 2" − L . Prove that L is a group under the operation æ restricted to L (such a subset is called a subgroup of K ). Proof . Closure is given. Associativity follows from the lemma at the beginning of this section's solutions. For each 2 − L , "K æ 2 œ 2 æ "K œ 2. Hence, "L œ "K. Finally, for each 2 − L, there is an 2" − L (by our assumption of closure of inverses) such that 22" œ 2 "2 œ "K œ "L . Hence, L is a group under the operation.
e
f
Problem 1.1.27 Prove that if B is an element of the group K then B8 l 8 − ™ is a subgroup of K (called the cyclic subgroup of K generated by B).
e
f
Proof . Call the set L œ B8 l 8 − ™ . Let 2ß 5 − L . Then there are :ß ; − ™ such that 2 œ B ; and 19). Since ; : − ™, 25 − L by definition. Therefore, L is 5 œ B: . Then 25 œ B; B: œ B ;: (exercise closed under the operation of KÞ Finally, if 2" œ B;, then 22 " œ B; B ; œ " and 2 "2 œ B ;B ; œ ". Since ; − ™, 2" − L by definition, so L is closed under inverses. By the previous exercise, L is a subgroup of K.
a b a a bca a bab bd ca a b a b a bca ba bd a aa b a ca b a bda
Problem 1.1.28 Let Eß æ group axioms for E ‚ F : .
b a b ba bda b a b ba b b b a b a b a a ba b a b
and Fß ˆ
be groups and let E ‚ F be their direct product. Verify all the
(a) prove that the associative law holds: for all +3 ß ,3 − E ‚ F , 3 œ "ß #ß $ , +" ß ," +#ß ,# +$ ß , $ œ + "ß , " + #ß , # + $ß , $ , (b) prove that "ß " is the identity of E ‚ F, and (c) prove that the inverse of +ß , is +" ß , " .
Proof . (a) Let +3 ß ,3 − E ‚ F with 3 œ "ß #ß $. Then
+" ß ,"
a a b a bb ba b b a b a b a b a b a ba b
+#ß ,# + $ß , $ œ + "ß , " + # æ + $ß , # ˆ , $ œ + "æ + #æ + $ ß , " ˆ , # ˆ , $
œ
+" æ + #
æ +$ ß
œ +" ß , " + # ß , #
, " ˆ ,# ˆ ,$ œ +"æ + #ß , " ˆ , # + $ß , $ + $ß , $ .
The intermediate step follows because +" æ +# æ +$ œ +" æ +# æ +$ and ," ˆ ,# ˆ ,$ œ ," ˆ ,# ˆ ,$ by the fact associativity holds for these elements because E and F is a group.
a b
(b) Let +ß , − E ‚ F . Then "ß " +ß , œ "æ +ß " ˆ , œ +ß , œ + æ "ß , ˆ " œ +ß , "ß " .
Robert Krzyzanowski
a
(c)
a b ba b
+ß , − E ‚ F. +ß , .
Let
"
+ ß,
"
Solutions to Dummit and Foote
Then
a ba
b a
Chapter 1
b a b a
b
+ß , +"ß , " œ + æ +" ß , ˆ , " œ "ß " œ + " æ +ß , " ˆ , œ
Problem 1.1.29 Prove that E ‚ F is an abelian group if and only if both E and F are abelian.
a b a b aa ba b b a b a ba b a b a ba b a b a b a b a ba b a b a b a ba b a b a b a b kk kk a ba b a b a b a ba b kk kk a b a a bb a b ˆ ‰ ˆ ‰ a a b b a b k k k a b k k k k a b k k k a b ˆ ‰ a b a b ˆ ‰ a b a b Šˆ a b ‰ a b ‹ e f a b kk a b k k a b Assume E ‚ F is abelian. Then for all +" ß +# − E and , "ß , # − F, it is true + "ß , " + #ß , # œ +# ß ,# +"ß ," . However, + "ß , " + #ß , # œ + "+ #ß , ", # and + #ß , # + "ß , " œ + #+ "ß , #, " . But then +" +# ß ,",# œ + #+ "ß , #, " , and the components must be equal by definition, so that + "+ # œ + #+ " and ," ,# œ ,# ," . Hence, E and F are abelian. Now assume this. Let +"ß , " ß + #ß , # − E ‚ F with + "ß + # − E and ," ß ,# − F. Then +" ß ," + #ß , # œ +"+ #ß , ", # œ + #+ "ß , #, " œ + #ß , # + "ß , " . The intermediate step is justified since we assumed E and F are abelian. By definition, we now know E ‚ F is abelian. Proof .
Problem 1.1.30 Prove that the elements +ß " and "ß , of E ‚ F commute and deduce the order of +ß , is the least common multiple of + and , .
Proof . We see +ß " "ß , œ + † "ß " † , œ " † +ß , † " œ "ß , +ß " . The intermediate step is justified because + − E, , − F, and E and F are groups so multiplying + or , by the identity is commutative. 8 8 Let + œ 8 and , œ 7. Then +ß " œ + 8ß " 8 (by a trivial inductive argument) and so +ß " œ "ß " . If 5 there were a 5 − ™ with 5 8 such that +ß " œ "ß " . then + 5 ß " 5 œ + 5 ß " œ "ß " so that + 5 œ ", contradicting the fact + œ 8. Hence, +ß " œ 8 œ + . Similarly, "ß , œ 7 œ , . Now let # # œ lcm 7ß 8 with # œ !7 and # œ " 8 for !ß " − ™ . Then +ß , œ + # ß , # œ + !7ß , " 8 œ
+7 ! ß , 8
"
+ß , œ "! ß "" œ "ß " . Now assume there is a $ − ™ such that $ # and
$
œ ". Then
+$ ß ,$ œ " so that +$ œ ,$ œ " . Assume 8 Îl $ so that $ œ :8 < for some :ß < − ™ ! with ! < 8. Then +$ œ + :8< œ + :8+ < œ + 8 : + < œ " :+ < œ + < œ " . However, this again contradicts the fact + œ 8 since < 8, so that 8 l $. Similarly, 7 l $. But then by definition $ lcm 7ß 8 œ #, a contradiction. Hence, +ß , œ lcm 7ß 8 .
Problem 1.1.31 Prove that any finite group K of even order contains an element of order # .
k k e f
Proof . Let K œ #8. If there was an element of order #, say B, we would have B# œ " so that B œ B ". Let L œ 1 − K l 1 Á 1" . Clearly, "  L . Furthermore, consider the following procedure. Let L! œ L and define L3" by removing some pair 13 ß 1" from L3 with 13 − L 3; that is L 3" œ L 3Ï 1ß 1 " . In each 3 iteration, we know 1 Á 1" so that two elements are removed. Since L is finite (as K is finite), eventually L5 œ g for some 5. But then L5" œ L5 # œ #, L 5# œ L 5" # œ %, etc., so that L! œ L œ #7 for some 7 − ™. Therefore, L has an even number of elements. Since "  L, L K . It is impossible that L œ K " since K 1 is odd. Hence, L K ". In other words, KÏ L " Á g. But this means there is some B − K such that B Á " with B œ B" (B  L by definition). That is, B œ #.
kk kk kk k k a k ek fb
kk k k kk
e f k k k k kk kk kk
k k k k kk kk
e f
Problem 1.1.32 If B is an element of finite order 8 in K, prove that the elements "ß Bß B# ß ÞÞÞß B 8" are all distinct. Deduce that B Ÿ K .
kk
Proof . Assume B3 œ B4 for some 3 Á 4 (! Ÿ 3ß 4 8 ). Without loss of generality, let 3 4. Then B34 œ !, but 3 4 8, so this contradicts the fact B œ 8. Hence, "ß Bß B #ß ÞÞÞß B 8" are all distinct. There are 8 of these elements, so K 8 œ B .
kk
Problem 1.1.33 Let B be an element of finite order 8 in K.
(a) Prove that if 8 is odd then B3 Á B3 for all 3 œ "ß #ß ÞÞÞß 8 "Þ
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
(b) Prove that if 8 œ #5 and " Ÿ 3 Ÿ 8 then B3 œ B3 if and only if 3 œ 5 . (a) If 8 œ " there is nothing to prove so assume 8 ". Assume B3 œ B3 for some " Ÿ 3 Ÿ 8 ". Then B3 B3 œ " so that B #3 œ ". Clearly, #3 Á 8 since #3 is even and 8 is odd. If #3 Ÿ 8 there would thus be a contradiction (since B œ 8 #3). Since 3 8, we then know 8 #3 #8 so that ! #3 8 8. However, B#3 œ B #38 8 œ B #38B 8 œ B #38 œ " . In other words, we found a positive integer less than 8 such that B to the power of that integer is ". But this contradicts the fact B œ 8. Hence, B3 Á B3 for all " Ÿ 3 Ÿ 8 ". Proof .
a
kb k
kk
kk
(b) Let B3 œ B3 for " Ÿ 3 Ÿ 8 and assume 3 Á 5. Then B 3B 3 œ " so that B #3 œ ". By assumption, #3 Á 8. If #3 Ÿ 8 there would thus be a contradiction (since B œ 8 #3). Since 3 8, we then know 8 #3 #8 so that ! #3 8 8. However, B#3 œ B #38 8 œ B #38B 8 œ B #38 œ " . In other words, we found a positive integer less than 8 such that B to the power of that integer is ". But this contradicts the fact B œ 8. Hence, 3 œ 5 . Now assume 3 œ 5 . Since " œ B8 œ B#5 œ B 5B 5 , we have B 5 œ B 5 , or B 3 œ B 3.
a b
kk
Problem 1.1.34 If B is an element of infinite order in K, prove that the elements B8 , 8 − ™ are all distinct.
Assume B3 œ B4 for some 3ß 4 − ™ with 3 Á 4. Then B 34 œ " . If 3 4 ! , this contradicts the " œ " " œ " so that B 34 œ B 43 œ ". Then 4 3 !, assumption B has infinite order. If 3 4 !, B34 but again this contradicts the assumption B has infinite order. Hence, B3 Á B4 for all 3ß 4 − ™ with 3 Á 4; that is, the elements B8, 8 − ™ are all distinct. .
a b
Proof .
a b
e
f
Problem 1.1.35 If B is an element of finite order 8 in K, use the Division Algorithm to show that any integral power of B equals one of the elements in the set "ß Bß B# ß ÞÞÞß B 8" (so these are all the distinct elements of the cyclic subgroup of K generated by B). Proof . Let 5 − ™. Then 5 œ ;8 < for some ; − ™, < − ™ with ! Ÿ < 8 by the Division Algorithm. Hence,
ab
B5 œ B;8< œ B;8B< œ B8 ; B< œ " ; B < œ B < with ! Ÿ < 8 as required.
e
f
Problem 1.1.36 Assume K œ "ß +ß ,ß - is a group of order % with identity ". Assume also that K has no elements of order %. Use the cancellation laws to show that there is a unique group table for K . Deduce that K is abelian. Proof . Assume there are two group tables Q" and Q # with the same rows and colums both "representing" "ß +ß ,ß and -, in that order (so that 1# œ +, 1$ œ ,, and 1% œ -). Now assume there is some 3ß 4 such that Q" B34 Á Q# B 34 where Q 5 B 34 is the 34th entry of group table (matrix) Q 5.
ab ab
ab
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Chapter 1.2 Problem 1.2.1 Compute the order of each of the elements in the following groups:
(a) H' (b) D)
(c) H"! .
Solution. This problem is trivial. The solution for each group is
kk k k k k kk kk k k k k kk k k k k k k kk kk k k k k k k kk k k k k k k k k
(a) The orders are l"l œ ", l
(b) The orders are " œ ", < œ <$ œ %, <# œ = œ =< œ =< # œ =< $ œ #.
(c) The orders are " œ ", < œ <# œ <$ œ <% œ &, = œ =< œ =< # œ =< $ œ =< % œ #. Problem 1.2.2 Use the generators and relations Ø<ß = l < 8 œ = # œ " , <= œ =< "Ù to show that if B is any element of H#8 , which is not a power of <, then
ab
a b a b
Proof . If B is not a power of <, B œ =<3 for some ! Ÿ 3 8. Then <= œ =< " so that <= < 3 œ =< "< 3, or in other words, < =<3 œ =< 3 < ".
Problem 1.2.3 Use the generators and relations above to show that every element of H#8 which is not a power of < has order #. Deduce that H#8 is generated by the two elements = and =< , both of which have order #.
a b aab b
If B is not a power of <, B œ =<3 for some ! Ÿ 3 8. But then < 3= œ < 3" <= œ < 3" =< " by 3 " one of the relations of H#8, and applying this 3 " times, <3 = œ = < " œ =< 3, so that < 3= œ < 3= and 3 hence < = œ #. Since = =< œ <, Ø=ß =<Ù œ Ø=ß =<ß <Ù œ Ø<ß =Ù œ H #8, H #8 is generated by = and =< .
a b
Proof .
k k
ab
Problem 1.2.5 If 8 is odd and 8 $, show that the identity is the only element of H#8 which commutes with all elements of H#8 . Proof . Let 8 be odd and greater than ". No <3 commutes with all elements for some " Ÿ 3 8. If it did, it would particularly for = giving <3 = œ =< 3, but our knowledge that < 3= œ =< 3 would then mean =<3 œ =<3 , or < 3 œ < 3 or < #3 œ ". Since " Ÿ 3 8, we know # Ÿ #3 #8 so that the former statement can only be true when #3 œ 8. The fact 8 is odd gives a contradiction. Finally, assume =<3 commutes with all elements for some " Ÿ 3 8. But then
a b a b
a b a b
a ba b a ba b
=<3 < œ < =< 3 Í =< 3" œ <=< 3 Í = =< 3" œ = <=< 3 Í < 3" œ =< =< 3 œ < "= =< 3 œ < 3"
a b a bœ<
which implies < 3"
3"
#
œ " again giving a contradiction.
kk
Problem 1.2.6 Let B and C be elements of order # is any group K . Prove that if > œ BC then >B œ B> " (so that if 8 œ BC _ then Bß > satisfy the same relations in K as =ß < do in H #8 ). " "
C B
ab ab
ab
The elements B and C are of order # so that B" œ B and C " œ C. Then > " œ BC œ CB so that >B œ BC B œ B CB œ B> " .
Proof .
ab
8
"
œ
Problem 1.2.7 Show that Ø +ß , l +# œ , # œ +, œ "Ù gives a presentation for H #8 in terms of the two generators + œ = and , œ =< of order # computed in Exercise $ above. Proof .
ab
Notice in the usual presentation of H#8, +, œ = =< œ <. Hence, we need to show that
Robert Krzyzanowski
Solutions to Dummit and Foote
ab
Ø +ß , l + # œ , # œ +,
8
ab
œ "Ù œ Ø +,ß + l +,
8
Chapter 1
ab ab
œ + # œ " , +, + œ + +,
ab
"
Ù
because the latter is the presentation of H#8 with +, œ < and + œ = (considered as literal strings, where the meaning of +, as "+ times ," can be dropped). Notice , œ +# , œ + +, . Hence, any element generated by + and , is also generated by + and +,, and conversely any element generated by +, and + is also generated by + and ,.
ab
8
To show the relations of the latter are implied by the former, notice we already have +, œ +# œ ". To " show the final relation, we need to prove +, + œ + +, but by the cancellation law this amounts to " # # ,+ œ +, . This is trivial since +, ,+ œ +, + œ + œ " so indeed ,+ œ +, ".
ab
a b a b a ba b a b ™
ab
Problem 1.2.8 Find the order of the cyclic subgroup of H#8 generated by < . Solution. We know
a ˜b
In each of Exercises 9 to 13 you can find the order of the group of rigid motions in ‘$ of the given Platonic solid by following the proof for order of H#8 : find the number of positions to which an adjacent pair of vertices can be sent. Alternatively, you can find the number of places to which a given face may be sent and, once a face is fixed, the number of positions to which a vertex on that face may be sent.
kk
Problem 1.2.9 Let K be the group of rigid motions in ‘$ of a tetrahedron. Show that K œ "#. Proof . A rigid motion of the tetrahedron is taking a copy of the tetrahedron, moving this copy in any fashion in 3-space, and then placing the copy back on the original 8-gon so it exactly covers it. Notice this rigid motion is a symmetry (and notice we could get more symmetries by moving through %-space, which can also be made into a group). We can describe these symmetries by choosing some labelling of the % vertices numerically from " to %. Then each symmetry = can be described uniquely by the corresponding permutation 5 À "ß #ß ÞÞÞß 8 Ä "ß #ß ÞÞÞß 8 where if the symmetry = places vertex 3 in the position where vertex 4 was originally, then 5 is the corresponding permutation sending 3 to 4. These symmetries can be made into a group K as follows. Define => for =ß > − K to be the symmetry obtained by applying > and then = to the tetrahedron. If =ß > effect the permutations 5ß 7 respectively on the vertices, then => will effect 5 ‰ 7. The binary operation of K is associative since composition of functions is associative. The identity of K is the identity symmetry (effecting the identity permutation), and the inverse of = − K is the symmetry that reverses all rigid motions of = (so that if = effects 5, =" effects 5 "). This shows we can indeed make a group out of the rigid motions of a tetrahedron, just like with H#8.
e
f e
f
kk
To show K œ "#, notice for any vertex 3 and some adjacent vertex 4, there is are $ symmetries which send vertex " into position 3. A particular symmetry is obtained by sending the vertex 4 to any of the $ adjacent vertices to 3. Hence, there are % ‚ $ œ "# possible such symmetries (since a tetrahedron has four vertices). These symmetries are the only symmetries of the tetrahedron, since a rigid motion in ‘$ is completely determined by where it positions three given points (apply this to the vertices 3, 4, and the unique vertex adjacent to these two).
kk
Problem 1.2.10 Let K be the group of rigid motions in ‘$ of a cube. Show that K œ #%.
kk
Proof . As in Problem 1.2.9, it can be verified that K is indeed a group. To show K œ #%, notice for any vertex 3 and some adjacent vertex 4, there is are $ symmetries which send vertex " into position 3. A
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
particular symmetry is obtained by sending the vertex 4 to any of the $ adjacent vertices to 3. Hence, there are ) ‚ $ œ #% possible such symmetries (since a cube has eight vertices). These symmetries are the only symmetries of the cube, since a rigid motion in ‘$ is completely determined by where it positions three given points (apply this to the vertices 3, 4, and the unique vertex adjacent to these two).
kk kk
Problem 1.2.11 Let K be the group of rigid motions in ‘$ of a octahedron. Show that K œ #%. Proof . As in Problem 1.2.9, it can be verified that K is indeed a group. To show K œ #%, notice for any vertex 3 and some adjacent vertex 4, there is are % symmetries which send vertex " into position 3. A particular symmetry is obtained by sending the vertex 4 to any of the % adjacent vertices to 3. Hence, there are ' ‚ % œ #% possible such symmetries (since an octahedron has six vertices). These symmetries are the only symmetries of the octahedron, since a rigid motion in ‘$ is completely determined by where it positions three given points (apply this to the vertices 3, 4, and the unique vertex adjacent to these two).
kk kk
Problem 1.2.12 Let K be the group of rigid motions in ‘$ of a dodecahedron. Show that K œ '!. Proof . As in Problem 1.2.9, it can be verified that K is indeed a group. To show K œ '!, notice for any vertex 3 and some adjacent vertex 4, there is are $ symmetries which send vertex " into position 3. A particular symmetry is obtained by sending the vertex 4 to any of the $ adjacent vertices to 3. Hence, there are #! ‚ $ œ '! possible such symmetries (since an octahedron has twenty vertices). These symmetries are the only symmetries of the dodecahedron, since a rigid motion in ‘$ is completely determined by where it positions three given points (apply this to the vertices 3, 4, and the unique vertex adjacent to these two).
kk kk
Problem 1.2.13 Let K be the group of rigid motions in ‘$ of an icosahedron. Show that K œ '!. Proof . As in Problem 1.2.9, it can be verified that K is indeed a group. To show K œ #%, notice for any vertex 3 and some adjacent vertex 4, there is are & symmetries which send vertex " into position 3. A particular symmetry is obtained by sending the vertex 4 to any of the & adjacent vertices to 3. Hence, there are "# ‚ & œ '! possible such symmetries (since an icosahedronhas twelve vertices). These symmetries are the only symmetries of the icosahedron, since a rigid motion in ‘$ is completely determined by where it positions three given points (apply this to the vertices 3, 4, and the unique vertex adjacent to these two).
Problem 1.2.14 Find a set of generators for ™.
ef
a b
Solution. Let the group operation be and an inverse of an element 5 − ™ be written 5. It suffices to consider " . First, notice ! œ " " . Furthermore, for 5 − ™ ,
a b a b a b ðóóóóóóóóóóóñóóóóóóóóóóóò
5 œ " " ÞÞÞ " , 5 times
and for 5 − ™ ,
ðóóóóñóóóóò
5 œ " " ÞÞÞ " . 5 times Hence, each element in ™ can be represented as a sum of "'s and "'s (the group operation applied to "'s and its inverse). By definition, " is a set of generators for ™, that is ™ œ Ø"Ù.
ef
Problem 1.2.15 Find a set of generators and relations for ™Î8™ .
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
ˆ ‰ ðóñóò ˜™
Solution. Let + − ™Î8™. Then ! œ " " and we for any other element
+ œ " ÞÞÞ " +w times
where +w − ™ with ! Ÿ + 8 so that + w œ +. Hence, " generates ™Î8™. If for 8 − ™ we define 8+ to be ! if 8 œ !, the sum + + ÞÞÞ + taken 8 times if 8 is positive, and the sum + " ÞÞÞ + " taken 8 times if 8 is negative, then the only relation we need for ™Î8™ is 8" œ !. Notice ! † "ß " † "ß ÞÞÞß 8 " † " must all be distinct elements so we know the order of a group with this generator and relation must be at least 8. Furthermore, these elements must be the only elements of such a group. For any element +, the group being generated by " means there is a 5 − ™ such that + œ 5". By the division algorithm, there are :ß ; − ™ with ! Ÿ ; 8 such that 5 œ :8 ; , so that + œ :8 ; " œ :8 " ;" œ : 8" ;" œ : † ! ;" œ ;" where 8" œ ! was given by our sole relation. In other words, any element + must be one of ! † "ß " † "ß ÞÞÞß 8 " † ". Hence, there are exactly 8 elements in such a group. In conclusion, the only such group is ™Î8™ so that
a b
ˆ‰
a b ab
a b
™Î8™ œ Ø" l 8" œ !Ù.
a b
Problem 1.2.16 Show that the group Ø B" ß C" l B#" œ C "# œ B "C " may be replaced by the letter < and C" by = ). Proof .
#
œ " Ù is the dihedral group H % (where B "
Recall the usual presentation of H#8,
Ø<ß = l < 8 œ = # œ " , <= œ =< "Ù . Then a presentation of H% œ H#†# is Ø <ß = l <# œ = # œ ", <= œ =< "Ù. If we replace B " by < and C " by =, we # want to show this presentation is equivalent to Ø <ß = l < # œ = # œ " , <= œ "Ù . The first relation in this # presentation is the same as the first in the usual presentation of H%, so we merely need to show <= œ " # implies the relation <= œ =<" . This is clear, since <= œ " means <=<= œ " so that <=<== "< " œ = "< " and on the left-hand side ==" cancel and then <<" cancel so we obtain <= œ ="< " œ =< " (since = # œ " gives = œ =" ).
ab
ab
ab
Problem 1.2.17 Let \ #8 be the group whose presentation is displayed in (1.2).
(a) Show that if 8 œ $5, then \#8 has order ' , and it has the same generators and relations as H' when B is replaced by < and C by = . (b)
a b
Show that if $ß 8 œ ", then B satisfies the additional relation: B œ ". In this case deduce that \#8 has order #.
Proof .
The presentation in question is given by
\#8 œ Ø Bß C l B8 œ C # œ "ß BC œ CB #Ù . (a) Assume 8 œ $5. As has been shown in the discussion in the book, \#8 is of order at most 'and a "hidden" relation is B$ œ ". Notice because 8 œ $5 the first relation B8 œ " can be replaced by B$ œ " since the former gives us no new information. Hence, B# œ B" so that BC œ CB ". In other words, given this knowledge about 8 we can rewrite the presentation,
\#8 œ Ø Bß C l B$ œ C # œ "ß BC œ CB "Ù . However, notice if B and C are replaced by < and = respectively, this is exactly the canonical presentation of H' . Therefore, a group satisfying the presentation for \#8 with at least ' elements exists, and so \#8 must be of order exactly '.
Robert Krzyzanowski
(b)
a b
Solutions to Dummit and Foote
Chapter 1
Assume $ß 8 œ " so that 8 œ $5 " for some 5 !. Then as in part (a) we know B$ œ ", so that
ab
5
B8 œ " gives " œ B8 œ B$5" œ B $ B " œ " 5B œ B. Hence, B œ " , and since any element must be of the form C5 B3 with ! Ÿ 3 Ÿ 8 " and 5 œ ! or ". This reduces to just C 5, so \ #8 has at most two possible elements, C and the identity. Additionally, the representation reduces to Ø C l C # œ "Ù. Notice from Problem 1.2.15, this is exactly the representation for ™Î#™ (with C# œ " written # † C œ !), so there is such a group \#8 with exactly # elements. Hence, this presentation must give a group with order #. Problem 1.2.18 Let ] be the group whose presentation is displayed in (1.3).
(a) Show that @# œ @" . (b) Show that @ commutes with ?$ . (c) Show that @ commutes with ?. (d) Show that ?@ œ ". (e) Show that ? œ ", deduce that @ œ ", and conclude that ] œ " . Proof .
Let ] be the group whose presentation is given by
] œ Ø ?ß @ l ? % œ @ $ œ "ß ?@ œ @ #? # Ù . Then since @$ œ ", we know @# @ œ " so that @# œ @ ". We now show @?$ œ ? $@. This is equivalent to showing @" ?$ @ œ ?$, or ?$ œ @ #? $@. This we can show because
a ba b a ba b ab a b a b a b a b
@# ?$ @ œ @#?# ?@ œ ?@ @ #? # œ ?@ $? # œ ? $. #
Hence @ commutes with ?$ . Next, notice ?* œ ?) ? œ ? % ? œ " #? œ ?. Then
a b a b
@? œ @?* œ @?$ ?' œ ? $@ ? ' œ ? $ @? $ ? $ œ ? $ ? $@ ? $ œ ? ' @? $ œ ? ' ? $@ œ ? *@ œ ?@ . Hence, @ also commutes with ?. Then the relation ?@ œ @# ?# implies @?@ œ ? # so that ?@ # œ ? # and so ?$ @# œ ", or ?"@ " œ " (since ?% œ " implues ? $ œ ? "). Hence, @? œ ?@ œ ". Then " œ ?@ $ œ ?$ @$ œ ?" so that ? œ ". Finally, the last relation is reduced to @ œ @ # so that the cancellation law gives @ œ ". Hence, the only element in this group is the identity, so it must be the trivial group; that is, ] œ ".
ab
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Chapter 1.3 Problem 1.3.1 Let 5 be the permutation
"È$
#È%
$È&
%È#
&È"
"È&
#È$
$È#
%È%
& È ".
and let 7 be the permutation
Find the cycle decompositions of each of the following permutations: 5 ß 7 ß 5 # ß 57 ß 75 ß and 7 #5 .
a ba b a ba b a ba b a ba b a b a ba b a ba b a b a ba b a ba b a b a ba b a ba b a ba b a ba b
Solution. First, notice 5 œ " $ & # % and 7 œ " & # $ . Hence,
5# œ 5 ‰ 5 œ " $ & # % ‰ " $ & # % œ " & $
57 œ 5 ‰ 7 œ " $ & # % ‰ " & # $ œ & $ % # . 75 œ 7 ‰ 5 œ " & # $ ‰ " $ & # % œ " # % $ .
7 #5 œ 7 ‰ 7 ‰ 5 œ " & # $ ‰ " & # $ ‰ " $ & # % œ " $ & # % . Problem 1.3.2 Let 5 be the permutation
" È "$ "" È (
# È # $ È "& % È "% & È "! ' È ' "# È * "$ È & "% È "" "& È )
( È "#
)È$
*È%
"! È "
* È "&
"! È $
and let 7 be the permutation
" È "% "" È )
# È * $ È "! "# È ( "$ È %
% È # & È "# ' È ' "% È " "& È "$.
(È&
) È ""
Find the cycle decomposition of the following permutations: 5 ß 7 ß 5 # ß 57 ß 75ß and 7 #5 . Solution. First, notice
a
ba ba b a ba ba ba ba b a ba ba ba b
a ba ba ba ba b a ba ba ba b a b
5 œ " "$ & "! $ "& ) % "% "" ( "# * and 7 œ " "% # * "& "$ % $ "! & "# ( ) "" . Hence,
5 # œ " & $ ) "& % "" "# ( * "% "! "$ 75 œ " % # * $ "$ "# "& "" & ) "! "% Notice 7 # 5 is one big cycle that fixes ' and *.
57 œ " "" $ # % & * ) ( "! "& "$ "%
7 # 5 œ " # "& ) $ % "% "" "# "$ ( & "!
Problem 1.3.3 For each of the permutations whose cycle decompositions were computed in the precesing two exercises, compute its order. Solution. The orders can either be computer directly (a very laborious process), or a quick mental inspection which will later be proved in Problem 1.3.15 yields
k k k kkk k k k k k k k k kk k k kk k k kk k k k k k k
5 œ "# , 7 œ $! ß 5 # œ 57 œ 57 œ ' , and 7 #5 œ "$
for the previous exercise and
5 œ 7 # 5 œ ', 7 œ #, 5 # œ 57 œ 75 œ %.
Robert Krzyzanowski
for the first exercise. Problem 1.3.4
Solutions to Dummit and Foote
Chapter 1
Compute the order of each of the elements in the following groups (a) W $ (b) W % .
Solution. As in the previous problem, in W $
k k k a bk bk ka bk ka bk bk ka bk ka bk bk ka bk ka ba bk ka ba bk ka
ka bk ka bk ka bk ka bk
" œ ", # $ œ " $ œ " $ œ #, " # $ œ " $ # œ $
and in W% the above permutations (the ones that fix %) all have the same orders, and
ka ka ka ka
" % œ # % œ $ % œ #,
ka bk ka bk ka bk bk ka bk ka bk ka bk ba bk
" # % œ # " % œ " $ % œ $ " % œ # $ % œ $ # % œ $,
" # $ % œ # " $ % œ " $ # % œ # $ " % œ $ " # % œ $ # " % œ %, and " # $ % œ " $ # % œ " % # $ œ #.
The author of the book probably intended these elements to be written the permutations to be written explicitly (e.g., define 53 for each of the 8x permutations in W 8), but finding the cycle decomposition is a fairly trivial process for these elements, so they were written as such, and hence they are indeed all unique elements. Note doing this makes Problems 1.3.6 and 1.3.7 trivial. Problem 1.3.5 Proof .
Problem 1.3.6 Proof .
Problem 1.3.7 Proof .
Problem 1.3.8
a
ba ba ba b ba ba ba bk
Find the order of " "# ) "! % # "$ & "" ( ' * .
ka
As in Problem 1.3.3, " "# ) "! % # "$ & "" ( ' *
œ $!.
Write out the cycle decomposition of each element of order % in W % .
See Problem 1.3.4. Write out the cycle decomposition of each element of order # in W % .
See Problem 1.3.4.
e
f
Prove that if H œ "ß #ß $ß ÞÞÞ then WH is an infinite group (do not say _x œ _ ).
ab
ef
ke fk k k
Proof . Consider permutations of the form 53 œ " 3 , that is, those for which " and 3 are exchanged and all other elements are fixed. There are an (countably) infinite such 53 and since 53 © WH, 53 Ÿ WH so that W H is an infinite group.
Problem 1.3.9
a
b
(a) Let 5 be the "#-cycle " # $ % & ' ( ) * "! "" "# . For which positive integers 3 is 5 3 also a "#-cycle?
a
b
(b) Let 7 be the )-cycle " # $ % & ' ( ) Þ For which positive integers 3 is 7 3 also an )-cycle?
a
b
(c) Let = be the "%-cycle " # $ % & ' ( ) * "! "" "# "$ "% Þ For which positive integers 3 is =3 also a "%cycle?
a b e
e
f
Proof . It is an obvious result (proved in Problem 1.3.11) that for an 8-cycle 5, 5 3 is an 8-cycle if and only if 8ß 3 œ ". Hence, (a) for a "#-cycle, 5 3 is also a "#-cycle if and only if 3 − "ß &ß (ß "" , (b) for an )cycle, 5 3 is also an )-cycle if and only if 3 − "ß $ß &ß ( , and (c) for a "%-cycle, 5 3 is also a "%-cycle if and only if 3 − "ß $ß &ß *ß ""ß "$ Þ
f
e
f
a
b
ek k f a b
Problem 1.3.10 Prove that if 5 is the 7-cycle +" +# ÞÞÞ + 7 , then for all 3 − "ß #ß ÞÞÞß 7 , 5 3 + 5 œ + 53 , where 5 3 is replaced by its least residue mod 7 when 5 3 7 . Deduce that 5 œ 7 .
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
ab ab a b a b a a bb a b a b a b e f ab a b ab ab ab a b kk ab kk a b ab e f aa b b a b e f a b a b a b a b a b a ba b a b a b ab a b a b a b a b a b ab a b a b
Proof . We proceed inductively. First, notice 5 +5 œ +5" where 5 " is considered mod 7. This is because if 5 7, then 5 assigns +5 to the element in the cycle adjacent on the right to +5, i.e., + 5". If 5 œ 7, then 5 assigns +5 to the first element in the cycle, i.e., +" œ + 7" œ + 5". Now assume that 5 3 +5 œ +53 for some 3 7. Then 5 3" + 5 œ 5 ‰ 5 3 + 5 œ 5 5 3 + 5 œ 5 + 53 . Again, if 5 3 7 , then 5 assigns +53 to the element in the cycle adjacent on the right to +53, i.e., + 53". If 5 3 œ 7, then 5 assigns +53 to the first element in the cycle, i.e., +" œ +7" œ + 53". Hence, 5 3" + 5 œ 5 + 53 œ +5 3" . Therefore, for any 3 − "ß #ß ÞÞÞß 7 , 53 + 5 œ + 53. Since
5 ! +5 œ + 5! ß 5 " + 5 œ + 5"ß ÞÞÞß 5 7" + 5 œ + 5
7"
are all distinct elements, 5 7. However, 5 7 +5 œ +57 œ + 5 then implies 5 œ 7.
Let 5 be the 7-cycle " # ÞÞÞ 7 . Show that 5 3 is also an 7 -cycle if and only if 3 is Problem 1.3.11 relatively prime to 7. Proof . Assume 3 Ÿ 7. By the previous exercise, 5 3 +5 œ + 53 and the inductive argument is easily extended to 3 7 by recognizing +5 for 5 − ™ define equivalence classes on + "ß ÞÞÞß + 7 with + 5 ´ + 3 for " Ÿ 3 Ÿ 7 if and only if 5 ´ 3 (mod 7). Hence, 583 +5 œ + 583. Notice 5 83 + 5 œ + 5 for all + 5 (i.e., 5 3 is by definition an 8-cycle) if and only if 5 ´ 5 83 mod 7 for all 5 − "ß #ß ÞÞÞß 7 which holds if and only if 83 ´ ! mod 7 . This is obvious for 3 œ 7 so assume 3 7. If 3ß 7 œ ", the lowest such integer 8 is 7 since any other integer 8 7 must satisfy 7 l 83 and hence also 7 l 3 (by the fact 8 is relatively prime to 7 and so they can have no common factors), which is impossible since 3 7. Hence 5 3 is an 7-cycle. The other direction follows when we show that if the lowest integer 8 such that 83 ´ ! mod 7 is 8 œ 7, then 3ß 7 œ ". Assume this hypothesis. Then if 3ß 7 Á ", i.e., if b. " such that . l 3 and . l 7, this would mean that 7Î.ß 3Î. − ™ so that 7Î. 3 œ 7 3Î. ´ ! mod 7 would contradict that 7 is the lowest such integer for 8 (since 7Î. 7). Hence 3ß 7 œ ".
To finish the argument for 3 7, notice by the division algorithm 3 œ ;7 : for ;ß : − ™, ! Ÿ : 7. Since 5 7 +5 œ +57 œ + 5 , we have 5 7 ´ ". Then ;
;
5 3 +5 œ 5 ;7: +5 œ 5 ;7 5 : +5 œ 5 7 5 : + 5 œ " 5 : + 5 œ 5 : + 5
for which we have already proven the statement.
Problem 1.3.12
a ba ba ba ba b a ba b a
a b
(a) If 7 œ " # $ % & ' ( ) * "! determine whether there is an 8-cycle 5 8 "! with 7 œ 5 5 for some integer 5 .
a b
(a) If 7 œ " # $ % & determine whether there is an 8-cycle 5 8 & with 7 œ 5 5 for some integer 5 .
b
Proof . (a) Yes, namely 5 œ " $ & ( * # % ' ) "! for 5 œ &. This can be deduced by realizing each of the elements in each #-cycle must be "!Î# œ & elements apart in a "!-cycle so that they can all "reach each other" while passing over the elements in the other #-cycles.
a
b
(b) No. We can attempt to construct such a 5. If there was such a 5 œ +" +# ÞÞÞ + 8 , then assume without loss of generality (we can cyclically permute the elements in the cycle until we get) +" œ ". If 8 &, then in order to fix +3 l +3 & through 5 5 (as 7 does), we would need 8 l 5. If this were false, take +3 to be such that +3 œ ', and then (decomposing 5 as usual with the division algorithm for ! < 8)
e
f
ab
a b a b a b ab a b a b e f ;
;
5 5 +3 œ 5 ;8< +3 œ 5 8 5 < + 3 œ " 5 < + 3 œ 5 < + 3 œ + 3< Á + 3 so that ' would not be fixed. However, then 8 l 5 would imply for any 3 − "ß ÞÞÞß 8 ,
Robert Krzyzanowski
Solutions to Dummit and Foote
a b a b a ba b a b a b a b a ba b ab ab 5 5 +3 œ 5 ;8 +3 œ 5 8
;
+3 œ " ; + 3 œ + 3 ,
a
that is, any element in the 8-cycle would be fixed. Hence, does not work either, since we need 5 5 +" œ 5 5 " œ # " œ +" , so that 55 ‰ 5 5 +" œ +", and hence 5 #5 +" #5 ´ ! mod & . However, #ß & œ " so we would
a b
Chapter 1
ba b a b a b ab
8 œ &, meaning 5 œ " +# +$ +% +& . However, this œ +3 (for some # Ÿ 3 Ÿ &) and 5 5 + 3 œ + 5 # œ œ + "#5 œ + " meaning " #5 ´ " mod & , or need & l 5. However, then 5 5 +" œ +"5 œ +", œ +3. Hence, there is no such 8-cycle 5 for 8 & (and indeed, none for
contradicting our assumption 5 5 +" any 8).
Problem 1.3.13 Show that an element has order # in W 8 if and only if its cycle decomposition is a product of commuting #-cycles.
a
ab
b
Proof . Assume an element 5 has order #Þ If its cycle decomposition had an <-cycle +" ÞÞÞ + < for < # we would attain a contradiction as follows. Since 5# +" œ +"# œ +$ Á + " (because this <-cycle has at least $ elements), 5 # would not fix +" so that by definition the order of 5 Á #. Hence its cycle decomposition is a product of commuting #-cycles.
a b a b a b a a bb a b
a b a a b b a b a b kk
Conversely, assume 5 œ +" ," ÞÞÞ + < , < . Then for " Ÿ 3 Ÿ <, 5 # + 3 œ 5 5 + 3 similarly 5 # ,3 œ 5 5 ,3 œ 5 +3 œ , 3 and hence 5 # ´ " so that 5 œ #.
œ 5 , 3 œ 5 + 3 and
Let : be a prime. Show that an element has order : in W 8 if and only if its cycle Problem 1.3.14 decomposition is a product of commuting : -cycles. Show by an explicit example that this need not be the case if : is not prime.
kk
Proof . Assume 5 − W8 is a product of commuting :-cycles 53, 5 œ : 53 œ : for each 3 so that 53 œ ". By commutativity of the cycles,
5 : œ
a# b # #
ak k b kk a b a b a b
53
:
œ
:
53 œ
#
53. By Problem 1.3.10, notice
" œ ".
kk
Notice the order of 5 5 œ = could not be less than :, because by the disjointness of the :-cycles this would imply that 53= œ " for some 3 with = : (in fact, for all 3), which is a contradiction. Hence 5 œ :.
a a bb
Conversely, let 5 œ :. Assume 5 has a cycle decomposition with an <-cycle 7 œ +" ÞÞÞ + < for < Á : (with < "). Then we need 5 : +" œ 7 : +" œ + ": œ + ". This would mean " : ´ " mod < , so that : ´ ! mod < . This is clearly false when < :, so consider < :. The condition holds when < l :, but this 5 has a cycle decomposition with would imply either < œ : or < œ ", a contradiction since " < :. Hence only :-cycles. Take the unique cycle decomposition and its : -cycles will be disjoint, so that they commute. This need not be true when : is not prime. The element commuting %-cycles.
a b
" $ # % − W% has order # but is a product of
Problem 1.3.15 Prove that the order of an element in W 8 equals the least common multiple of the lengths of the cycles in its cycle decomposition.
#
kk a b ab a b ab ab kk f
Proof . Assume the order of an element 5 − W8 is smaller, with 5 œ 3 53 for 53 disjoint 83-cycles. Then there is such an 3 so that 83 œ l53 l Îl 8 ³ 5 (the former equality is by Problem 1.3.10), meaning 8 œ ;83 < for ;ß < − ™ with ! < 8 3 . First, notice it is necessary 538 œ 5 8 œ 5383 œ " (since all elements fixed by 5 8 must be fixed by 538, and also the order of 53 is 83). If 53 œ + " ÞÞÞ + 83 , then
ab e kk
;8 3<
538 +" œ 53
; +" œ 5383 5 < +" œ 5< +" œ + "< Á + "
since ! < =3. This contradicts that 5 œ = (since one of its cycles does not fix + " when applied to it = times). If 8 œ lcm 83 œ 53 À 53 a 8 3-cycle in the cycle decomposition of 5 , then 8 3 l 8 (say 8 œ ; 38 3) so
Robert Krzyzanowski
kk
Solutions to Dummit and Foote
58 œ
a# b # # 3
53
8
œ
8 3 53
œ
3 53
;3 83
œ
#a b #a b 3
5383
;3
œ
3
"
Chapter 1 ;3
œ ".
Hence 5 œ 8. Problem 1.3.16 Show that if 8 7 then the number of 7-cycles in W8 is given by
a
a ba b a
b.
8 8" 8# ÞÞÞ 87" 7
b
a ba b a
b
Proof . Let 5 œ +" ÞÞÞ + 7 . There are 8 ways of choosing + ", 8 " ways of choosing + # (since + " can not appear again), etc., so there are in total 8 8 " 8 # ÞÞÞ 8 7 " possible 7-cycles. However, not all of these are unique, since cyclically permuting the elements of an 7-cycle is an equivalence relation. For each 7-cycle, there are 7 such permutations, corresponding to +3 È +3, ÞÞÞß + 3 È + 37 (that is, + 3 È + 35 means the cycle's elements are shifted to the right by 5 spots, leaving the same cycle but a different representation). Hence, there are in total
a ba b a
b
8 8" 8# ÞÞÞ 87" 7
7-cycles in W 8.
a ba ba b a ba b a b e f e f ke f e fk ke fk ke fk a b a b ab a b a b a ba b a ba ba b a ba b a ba b a ba b
Problem 1.3.17 Show that if 8 % then the number of permutations in W8 which are the product of two disjoint #-cycles is n 8 " 8 # 8 $ Î).
The elements in question are of the form + , - . . By the previous problem, there are 8 8 " Î# ways to form the first #-cycle. However, now we need -ß . − "ß ÞÞÞß 8 +ß , to satisfy that we have disjoint cycles. Since "ß ÞÞÞß 8 +ß , œ "ß ÞÞÞß 8 +ß , œ 8 #, there are 8 # ways of picking -, and then 8 $ ways of picking . so that - Á .. Hence, there are 8 # 8 $ ways in total. However, if - . is a 2-cycle, . - is equivalent to it so we divide by #. In summary, there are: Proof .
8 8" #
8# 8$ #
œ
8 8" 8# 8$ %
different elements which are products of disjoint #-cycles. However, when + , - . is a #-cycle, by commutativity of disjoint cycles + , - . œ - . + , . Hence, we have to divide by # again to obtain the final number of unique products of two disjoint #-cycles:
a ba ba b Î# œ a ba ba b .
8 8" 8# 8$ %
8 8" 8# 8$ )
Problem 1.3.18 Find all numbers 8 such that W& contains an element of order 8. Proof . The possible cycles in W& are #ß $ß %ß and & cycles. Hence, unique permutations in W& can be written as one of these, or a product of #- and #- or #- and $-cycles (no #- and %-, #- and &-, or %- and &cycles, since these would not be disjoint and hence not a valid cycle decomposition). Problem 1.3.15 gives
e
a b a bf e
8 − "ß #ß $ß %ß &ß lcm #ß # ß lcm #ß $
f
œ "ß #ß $ß %ß &ß ' .
Problem 1.3.19 Find all numbers 8 such that W( contains an element of order 8. Proof . The possible cycles in W( are #ß $ß %, &, ', and ( cycles. Hence, unique permutations in W( can be written as one of these, or a product of #- and #- ß #- and $-, 2- and 4-, 2- and 5-, $- and $-, $- and %-, cycles (none of the other products of cycles, since these could not be disjoint and hence not a valid cycle decomposition). Thus, by Problem 1.3.15,
e
a b a b a b a b a b a bf
e
f
8 − "ß #ß $ß %ß &ß 'ß (ß lcm #ß # ß lcm #ß $ ß lcm #ß % ß lcm #ß & ß lcm $ß $ ß lcm $ß % œ "ß #ß $ß %ß &ß 'ß (ß "!ß "# . Problem 1.3.20 Find a set of generators and relations for W $ .
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Proof . Notice a single generator would have to be of order '. However, by Problem 1.3.15 the only way to achieve this would be to have an element have a cycle decomposition of a 2- and a 3-cycle. However, this is impossible since that would require & distinct elements in "ß #ß $ , and there only are $. Hence, W$ is not generated by one element. Additionally, we cannot have commutative (disjoint) generators. Let 5 œ " # , 7 œ " $ . Then 5 # œ ", 75 œ " # $ , 57 œ " $ # , and 575 œ # $ , so indeed the elements $ generate W$. Then the relations 5 # œ 7 # œ 57 œ " suffice. It's easy to see the last relation implies # 57 œ 75 , which means 57 Á 75 and neither 57 nor 75 reduce to "ß 5 ß or 7 (by examining orders). This these would gives us five elements, "ß 5ß 7 ß 57 ß 75. Furthermore, notice no element can have a 5 # or 7 # since # just cancel out. Hence each element has to be a "chain" of single 5's and 7's. Furthermore, 57 œ 75 gives 575 œ 757 so that any chain of 5 and 7 's of length greater than $ can be reduced to a chain of length $ or less. Hence the only chain of length $ is 57 5, and these are all possible elements since any others can be # reduced. To verify this last one is unique, notice 57 5 œ ", but it cannot be either of the other two elements of order # (5 or 7 ), because 575 5 œ 57 Á " and 575 7 œ 757 7 œ 75 Á ". Hence, the elements "ß 5ß 7 ß 57 ß 7 5ß 57 5 are all unique and are the only possible elements, so the group is indeed of order '. Of course we constructed this presentation explicitly from W$, W $ satisfies these generators and relations:
ab
ab ab
a a bb a b
e f
ab
ab
a b
a ba b a b ab
W$ œ Ø +ß , l + # œ , # œ +,
$
œ "Ù.
Notice omitting the last relation would give us the infinite dihedral group Dih _ (the group of symmetries generated by a reflection and rotation, where the rotation is not a rational multiple of a full rotation; i.e., the group of symmetries of a circle), where +, is a product of two elements of finite order (with torsion) but itself is of infinite order (torsion-free).
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1.4
k a bk
Problem 1.4.1 Prove that KP# …#
Chapter 1
œ '.
Πab a b a b a b a b b k a bk ab
+ , with each +ß ,ß -ß . equal to 0 or ". This matrix is - . invertible (in KP# …# ) if and only if det E œ +. ,- Á !. That is, we need +. Á ,-. Notice if BC œ " if and only if B œ C œ " for Bß C − …#Þ Hence, if + œ " and . œ ", then either -ß , œ !ß ! , !ß " , or "ß ! . On the other hand, if +ß . œ !ß ! , !ß " , or "ß ! we would need -ß . œ "ß " . Hence we have six such possible +ß ,ß -ß . , so that KP# …# œ '. Proof .
a
Elements must be of the form E œ
a b a b a b a b a b a b
Problem 1.4.2 Write out all the elements of KP# …# and compute the order of each element. Solution.
ºŒ º ºŒ º ºŒ º ºŒ º ab " ! ! "
" " ! "
œ ",
œ
" ! " "
! " " "
œ
œ #, and
ºŒ º ºŒ º " " " !
œ
! " " "
œ $.
Problem 1.4.3 Show that KP# …# is non-abelian. Proof .
ΠΠΠΠΠΠ" " ! "
" ! " "
œ
! " " "
" " " !
Á
œ
" ! " "
" " . ! "
Problem 1.4.4 Show that if 8 is not prime then ™Î8™ is not a field.
e f a b
aa b b ‡
Proof . Assume 8 œ +, for some +ß , − ™ " . Then ™Î8™ ß † ‡ operation is not well-defined: + † , œ !  ™Î8™ . Hence it is not a field.
ab ab
is not a group because the
Problem 1.4.5 Show that KP8 J is a finite group if and only if J has a finite number of elements. Proof .
Assume KP8 J is finite and J is not. This is a contradiction because for each + − J,
Œ
+ !
! +"
ab
has non-zero determinant ( œ "), and each of these elements is distinct, so that KP8 J would have infinite elements.
ab
be of the form Take J to have finitely many elements, say <. Then E − KP8 J must
Πab k a bk a bk k ab + -
, .
with +ß ,ß-ß . − J ,
and there are hence at most <% distinct elements in KP 8 J ( < choices for + ‚ < choices for , ‚ ...) .
kk
Problem 1.4.6 If J œ ; is finite prove that KP8 J
#
;8 .
Proof . An element in KP8 J is a matrix with 8 rows and 8 columns, with each spot in the matrix an upper bound for the taking one of ; possible values (since J œ ; and the element must be in J). Hence, unique number of distinct elements in KP8 J is (taking +34 − J to be the 3th row and 4th column):
Robert Krzyzanowski
Solutions to Dummit and Foote
# #a 8
k a bk
To show KP8 J
8
b
Chapter 1
#
; choices for +34 œ ; 8 .
3œ" 4œ"
ab
#
; 8 it then suffices to find one element of this form not in KP8 J , e.g.,
Œ
! ! . ! !
ab
Problem 1.4.7 Let : be a prime. Prove that the order of KP# …: is :% : $ : # : .
ab
Proof . As in Problem 1.4.5, KP# …: has at most : % elements (all possible matrices, invertible and non-invertible). From linear algebra, we know a matrix is invertible if and only if its rows are linearly independent. Hence, matrices of the following form are NOT invertible:
Œ
+ , -+ -,
Πand
-+ -, + ,
(1)
Notice the second form of matrix can almost always be constructed by the first. For example, if we want to start with .ß / in the bottom row and look at multiples of it for the first row, we could let + œ -. and , œ -/ so that
Œ
-. .
-/ /
Œ œ
+
,
-" + - "/
.
However, notice - has a multiplicative inverse if and only if - − …‡: if and only if - Á !. Hence, the second form of matrix in (1) can be expressible in the first if and only if - Á !. This gives us a classification of all non-invertible matrices in KP# …: : elements are of the form
Œ
+ , -+ -,
ab Πor
! .
! /
with +ß ,ß -ß .ß / − J and + and , not both !.
Now it remains to count how many possible options there are. There are :# " choices for +and , (since + and , can be anything from J , except for both !). Once + and , are fixed, there are : choices for -. Hence, the number of matrices of the first form is :# " : œ : $ : . For the second form, there are no restrictions on . or /, so the number of choices is :#. Hence, in total, the number of non-invertible matrices in KP# …: is :$ : : # œ : $ : # : . Finally, a matrix is either invertible or not, so if Q # J is the set of # ‚ # matrices with entries in J ,
a b
k a bk k a bk k e KP# …:
ab
œ Q# …: E − Q# …: À E not invertible
ab a b a b! a b
fk
a
b
ab
ab
œ :% :$ :# : œ :% :$ :# :.
Problem 1.4.8 Show that KP8 J is non-abelian for any 8 # and any J .
!
Proof . A simple but not rigorous way of showing this is to consider just the outer edges of a matrix. 8 Assume E œ +34 ß F œ , 34 − KP 8 J . Then the top-left value for EF would be 5œ" + "5 , 5" and the top8 left value for FE would be 5œ" +5","5 . These do not have to be equal, although this is not immediately obvious, since the condition that E and F have an inverse could place an important restriction on +34 and ,34 so that these two sums are indeed equal.
a b a b
a b
To see a more specific example of this, let ! œ " J ß and " œ " J ߆ . That is, let !and " denote the additive and multiplicative identity in J , respectively. Finally, let " denote the additive inverse of ". Consider the matrix given by E œ +34 such that + 34 œ $3 84" , where $ is the Kronecker delta (i.e., let E
ab
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
ab
be the matrix with " in the bottom-left to top-right diagonal); and consider the matrix F œ ,34 with ,34 œ " $ 34 except for ,88 œ " (i.e., F is the matrix with "'s everywhere except the main diagonal, with the additional exception that the bottom-right element is "). Then notice that det E œ „ " (depending on 8) and 8+1 similarly det F œ " (the zero's in the diagonal mean the determinant is just the appropriate sign times the bottom-right element, "). Hence Eß F − KP8 J . Finally, it's then simple to show that EF is F mirrored vertically and FE is F mirrored horizontally. These are different matrices (the former has a "in the lower left corner whereas the latter does not), so KP8 J is not abelian.
a b
ab ab
Problem 1.4.9 Prove that the binary operation of matrix multiplication of # ‚ # matrices with real number entries is associative.
ab
G œ -34
ab e f a b aa ba bba b Π! a b Π! Π! a b a baa ba bb a bΠ! Π! Π!
ab ab
We can prove this in general for 8 " and elements in any field J. Let E œ +34 ß F œ , 34 ß − KP8 ‘ with +34ß , 34ß - 34 − J for 3ß 4 − "ß ÞÞÞß 8 . Then component-wise:
Proof .
8
EF G œ
+34 ,34
- 34 œ
+ 35 , 54
8
8
2œ"
5œ"
- 34 œ
+ 35, 52 - 24 , and
5œ"
8
E FG œ +34
,34 -34
œ + 34
8
, 32- 24
œ
2œ"
8
+ 35
5œ"
, 52- 24
.
2œ"
Notice these last two are equivalent, since we can rearrange terms due to the abelian nature of addition and multiplication in J (this is what gives associativity). Problem 1.4.10 Let K œ
œŒ
(a) Compute the product of
+ !
Œ
+" !
(b) Find the matrix inverse of
, l +ß ,ß - − ‘ß + Á !ß - Á ! . ," -"
Πand
Π+ !
, -
+# !
,# -#
to show that K is closed under matrix multiplication.
and deduce that K is closed under inverses,
ab
(c) Deduce that K is a subgroup of KP# ‘ . Solution.
Œ
(a)
+" !
," -"
Œ
+# !
,# -#
ΠΠ+ !
(b)
, -
Œ œ
+" +# !
"Î+ ,Î+! "Î-
Œ +" , # , "- # -" -# œ
− K.
" ! . ! "
Notice the inverse is in K (since + Á !ß - Á ! means all the entries are real) so K is closed under inverses. (c) From Problem 1.4.9, K is associative since it is a subset of # ‚ # matrices with real number entries. Additionally, "! !" is an identity for K, so that all the group axioms are satisfied. Then K is a group, and it is a subset of KP# ‘ , so it must be a subgroup.
ˆ‰
ab
The next exercise introuces the Heisenberg group over the field J and develops some of its basic properties. When J œ ‘ this group plays an important role in quantum mechanics and signal theory by giving a group theoretic interepretation (due to H. Weyl) of Heisenberg's Uncertainty Principle. Note also that the Heisenberg group may be defined more generally -- for example, with entries in ™ .
Robert Krzyzanowski
Problem 1.4.11
\œ
Î Ï
" + ! " ! !
, "
Ñ Ò
Solutions to Dummit and Foote
ÚÎ Ñ ÛÜÏ Ò Î Ñ Ï Ò
" + ! " Let L œ ! ! " . / ! " 0 and ] œ ! ! "
, "
Þ ß à ab aa bb a b aa b b Ñ Î Ò Ï
l +ß ,ß - − J
Chapter 1
-- called the Heisenberg group over J . Let
be elements of L J .
(a) Compute the matrix product \] and deduce that L J is closed under matrix multiplication. Exhibit explicit matrices such that \] Á ] \ (so that L J is always non-abelian ).
ab
(b) Find an explicit formula for the matrix inverse \ " and deduce that L J is closed under inverses.
ab
(c) Prove the associative law for L J
kk
$
and deduce that L J is a group of order J .
(d) Find the order of each element of the finite group L ™Î#™ .
(e) Prove that every nonidentity element of the group L ‘ has infinite order. Solution. (a)
\] œ
Î Ï
" + ! " ! !
, "
ÑÎ ÒÏ
" . ! " ! !
/ 0 "
" +. ! " ! !
œ
/ +0 , -0 "
Ñ Ò
.
Since J is closed under and † , + .ß / +0 ,ß - 0 − J so that \] − L . (b)
Î Ï
" ! !
+ , " ! ! " - ! " ! ! " ! ! "
»
Ñ Î Ò Ï œ
" ! ! " ! !
, +"
»
" + ! " ! !
Ñ Î Ò Ï
! ! "
œ
" ! ! " ! " ! ! ! ! " !
»
+ " !
+- , "
Ñ Ò
.
Hence,
\
"
ab ÑÑÎ ÒÒÏ
œ
Î Ï
" + +- , ! " ! ! "
Ñ Ò
−K
since +ß +- ,ß - − J . Hence, L J is closed under inverses. (c) This follows Problem 1.4.9. However, it is possible to do it explicitly:
ÎÎ ÏÏ œ
œ
" + ! " ! !
, "
Î a b Ï " ! !
Î Ï
ÑÎ ÒÏ
" . ! " ! !
/ 0 "
" 1 2 ! " 3 ! ! "
a a b ba
Ñ Î Ò Ï œ
bÑ Î Ò Ï
+ . 1 2 + . 3 / +0 , " -0 3 ! "
" + ! " ! !
, "
ÑÎ ÒÏ
" . 1 2 .3 / ! " 0 3 ! ! "
Ñ Î Ò Ï œ
" +. ! " ! !
œ
" + ! " ! !
/ +0 , -0 "
a ba
" + .1 ! " ! !
, "
ÑÎÎ ÒÏÏ
" . ! " ! !
ÑÎ ÒÏ
" 1 2 ! " 3 ! ! "
Ñ Ò
ba a b b Ñ Ò
2 .3 / + 0 3 , - 0 3 "
/ 0 "
ÑÎ ÒÏ
" 1 2 ! " 3 ! ! "
ÑÑ ÒÒ
.
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
(d) First, notice #
E œ
Î Ï
" #+ +- #, ! " #! ! " E% œ
Î Ï
Ñ Î Ò Ï œ
" ! +! " ! ! ! "
" %+ '+- %, ! " %! ! "
Ñ Ò
Ñ Ò
, and
œ ".
Hence each non-identity element has order between # and %. If +- œ ! (6 possible matrices), the matrix has order #. Otherwise, it's easily checked the cases +ß - œ ", , œ ! or " yield a matrix of order %. (e) We will show inductively that
E8 œ
Î Ï
" ! !
8+ " !
a b +- 8, Ñ
8 8" #
8"
Ò
.
This is trivial for 8 œ ". Assume E5 has this form for some 5 ". Then 5"
E
5
œE Eœ
œ
Î Ï
Î Ï
" ! !
5+ " !
a b +- 5, ÑÎ "
5 5" #
5"
a
Ñ ÒÏ Ò b Ñ Î a b Ò Ï + ! " ! !
, "
5 5"
" 5+ + , 5+- # +- 5, ! " 5- ! ! "
œ
" ! !
5" + " !
a b +- a5 "b, Ñ
5 5" #
a b Ò 5" "
.
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Chapter 1.5 Problem 1.5.1 Compute the order of each of the elements in U) . Solution. The orders of " and " are " and #, respectively, and everything else has order %.
Problem 1.5.2 Write out the group tables for W$ ß H) and U) . Solution.
Group table for W $:
a b aa bb aa bb a b a W$ ß ‰ " "# "$ #$ "#$ "$#
Group table for H):
H) ß ‰ " < <# <$ = =< =<# =<$
Group table for U):
bb aa bb b aa bb bb a b b b
a b U) ß † " " 3 3 4 4 5 5
aa aa aa
" " "# "$ #$ "#$ "$#
"# "# " "$# "$# #$ "$
aa a aa a
bb b bb b
"$ "$ "#$ " "$# "# #$
aa aa aa
bb bb bb
#$ #$ "$# "#$ " "$ "#
aa aa aa
bb bb bb
"#$ "#$ "$ #$ "# "$# "
aa aa a a
bb bb b b
"$# "$# #$ "# "$ " "#$
" " < <# <$ = =< =<# =<$
< < <# <$ " =< =< # =< $ =
<# <# <$ " < =< # =< $ " =<
<$ <$ " < <# =< $ = =< =< #
= = =< $ =< # =< " <$ <# <
=< =< = =< $ =< # < " <$ <#
=< # =< # <= = =< $ <# < " <$
=< $ =< $ <= # <= = <$ <# < "
" " " 3 3 4 4 5 5
" " " 3 3 4 4 5 5
3 3 3 " " 5 5 4 4
3 3 3 " " 5 5 4 4
4 4 4 5 5 " " 3 3
4 4 4 5 5 " " 3 3
5 5 5 4 4 3 3 " "
5 5 5 4 -* 4 3 3 " "
Problem 1.5.3 Find a set of generators and relations for U) Proof .
Notice we need two generators. Take Bß C. We will show necessary relations are
B# œ C # and CBC " œ B ".
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
If B œ 3ß C œ 4, and BC œ 5 , it's easy to see U ) is generated by and satisfies these relations. Namely, " œ 3 %, " œ 3# , 3 œ 3$ , 4 œ 4 $ , 5 œ 43 $ and 5 œ 34 $ . A group of order ) satisfying this presentation exists, so it must be of order 8 at least. [Stop reading here. Need to finish this.]
ab
ab ab ab ab
ab
$
Notice "ß 3ß 3#ß 3$ must all be different elements. Additionally, from the third relation 34 œ 4 $3 $ so that 43 % œ ", and also 34 #34 œ 4 $3 $ so that 34 # œ 34 4 #3 # 34 . But this gives 4 #3 # œ " , and from the first two relations 3# 4# œ " and 3 # œ 4 #. Additionally, 4 #3 # œ " gives us how to commute 3 and 4, namel
ab
ab
Hence, all powers of 4 greater than " can be expressed in 3, so only 4 itself can be unique. To verify it is, we see 4 œ 3< for some ! Ÿ < % gives a contradiction: if < were odd the fact 3#4 # œ " would give 3 5 œ " for a " 5 %, a contradiction; if < were even, 3# œ 4# œ " would then give a contradiction as well. Again, 43 œ 3< , 43# œ 3 < or 43 $ œ 3 < would give a similar contradiction, so that 4ß 43ß 43 #, and 43 $ are all distinct elements. Adding any further 3 or 4's to "ß 3ß 3#ß 3$ß 4ß 43ß 43 #ß or 43 $ will give use a reduction using our relations, so these are the only possible elements. Hence,
ab
U) œ Ø3ß 4 l 3 % œ 4 % œ 34
%
œ "Ù .
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Chapter 1.6 Problem 1.6.1 Let : À K Ä L be a homomorphism.
a b ab
(a) Prove that : B8 œ : B
8
for all 8 − ™ .
a b ab ˆ ‰ ab ˆ ‰ ˆ ‰ ˆ ‰ab ab ab ab a b a b a b a b a b a b a b ab a b a b ab a b ab a b a b ab a b ˆa b ‰ a b ˆ a b ‰ a b k a bk k k k a bk ab ab ab a b ab ab a b a b a b a b a b a ab b a b a b k k k a bk k a b k k k a b k k k k ˆ ‰
(b) Do part (a) for 8 œ " and deduce that : B8 œ : B Proof .
8
for all 8 − ™.
(a) This is a simple inductive argument. Assume : B5 œ : B
5
for some 5 − ™. Then
5
5"
: B5" œ : B 5 B œ : B5 : B œ : B : B œ : B
.
!
(b) First, notice : B! œ : " œ : B œ ". This is because : " œ : " † " œ : " : " . Furthermore, " œ : " œ : B" B œ : B " : B œ : BB " œ : B : B " , so that : B " œ : B " . Hence, for 8 − ™ ,
: B8 œ : B"
so that the statement holds for all 8 − ™.
8
œ : B "
8
œ :B
" 8
œ: B
8
,
Problem 1.6.2 If : À K Ä L is an isomorphism, prove that : B œ B for all B − K . Deduce that any two isomorphic groups have the same number of elements of order 8 for each 8 − ™ . Is the result true if : is only assumed to be a homomorphism? Proof .
Let : B
œ 8 − ™ so that : B
8
œ ". Of course : " œ " since
: " œ: "†" œ: " : " 8
8
so by cancellation : " œ ". Then : B œ : " . However, : B œ : B8 from the previous exercise. Hence, : B8 œ : " and since isomorphisms are injective, B8 œ ". Assume B was smaller, say 7 8. 7 But then : B7 œ " so that : B œ ", giving : B Ÿ 7 8, a contradiction. Hence : B œ B . Let K8 and L8 be the subsets of K and L with order 8. Since : is an isomorphism, : B − L 8 is unique for each B − K8. Hence, K8 œ L 8 , that is, two isomorphic groups have the same number of elements of order 8 for each 8 − ™. This result is not true if : is a homomorphism. For example, if : is prime then
: À ™Î:™ ‚ ™Î:™ Ä ™Î:™ , +ß , È +
is a homomorphism such that ™Î:™ ‚ ™Î:™ has :# " elements of order : whereas ™Î:™ only has : ". If Problem 1.6.3 If : À K Ä L is an isomorphism, prove that K is abelian if and only if L is abelian. : À K Ä L is a homomorphism, what additional conditions on : (if any) are sufficient to ensure that if K is abelian, then so is L ?
aa b b a b a ba b aa bb a b a ba b a b a b a b a ba b a b a b ef ef
ab a b a b aa b b
Proof . Assume K is abelian. If +ß , − L , there are unique Bß C − K such that + œ : B and C œ : , . Then +, œ : B : C œ : BC œ : CB œ : C : B œ ,+ . Assume L is abelian. If Bß C − K , then : BC œ : B : C œ : C : B œ : CB so that BC œ CB by the fact : is injective. Hence K is abelian if and only if L is. If : is a homomorphism and K is abelian, then for Bß C − K, : B : C œ : BC œ : CB œ : C : B . Hence, for two elements +ß , − L , if there are Bß C − K such that + œ : B and , œ : C , then we can say the two elements commute. In other words, : has to be surjective (but not necessarily injective).
Problem 1.6.4 Prove that multiplicative groups ‘ ! and ‚ ! are not isomorphic.
Robert Krzyzanowski
Solutions to Dummit and Foote
ef
ef
Chapter 1
e fk k
Proof . The group ‘ ! has no elements of order %, but ‚ ! has two ( „ 3), so by Problem 1.6.2 the groups are not isomorphic. To verify the former rigorously, notice if B − ‘ ! with finite order 5, then B5 œ " and B 5 œ B5 œ " (where † denotes absolute value, not order). But then B œ " (since the only non-negative real 5-th root of unity is "), so that either B œ " which has order ", or B œ " which has order #.
kk ¸ ¸
kk
Problem 1.6.5 Prove that the additive groups ‘ and are not isomorphic. Proof . There is no injection from ‘ to since the former is uncountable and the latter is countable. This can be seen more explicitly in the following argument. Let : À ‘ Ä be an isomorphism. First, notice if B − ‘ , then ! œ : ! œ : B B œ : B : B so that : B and : B are inverses in ß . Assume : " œ +, − . From the previous result, we can consider +ß , ! without loss of generality. Then : , œ ,: " œ +. Thus, : , œ : + † +, œ + † : +, œ + † " . Then : +, œ " (otherwise we would get + † - œ + for some - Á " in ß ). Let .- − be a lowest form representation. Then , -, .,,,: -, + œ - : + œ - , so that - œ : + œ : +. œ .: +. . This implies : +. œ . (otherwise, .B œ ,œ .- . This applies for any rational in , for some B Á .- in ß ). From earlier, this implies : +. so that : must map all rationals to rationals ( : l œ ). However, this contradicts the fact : is injective, since any irrational in ‘ must also map to a rational.
a b a b a b a b a b a b a b ˆ‰ ab ab a b aˆ ‰b ˆ ‰ ˆ ‰ ˆ ‰ a b ˆ ‰ ˆ ‰ ˆˆ ‰ ‰ ˆ‰
a b
Problem 1.6.6 Prove that the additive groups ™ and are not isomorphic.
aa b b a a b b a ba b a b
a b a b
a
ab
b
Let : À ™ Ä be an isomorphism. First, notice if B − ™, then ! œ : B B œ : B : B so that : B and : B are inverses in ß . Assume : " œ +, − . Then for 5 − ™ , +5 : 5 œ : 5 † " œ 5: " œ +5 , . From earlier, : 5 œ , . Hence, all elements in ™ must be mapped to + integer multiples of +, . However, then no element will be mapped to, e.g., #, . This contradicts the fact : is surjective. Proof .
Problem 1.6.7 Prove that H) and U) are not isomorphic. Proof . The group H) only has two elements of order % ( < and <$), whereas U) has six. By Problem 1.6.2, they cannot be isomorphic.
Problem 1.6.8 Prove that if 8 Á 7, W8 and W7 are not isomorphic.
kk k k
Proof . This is obvious, since then 8x Á 7x so that W8 Á W7 , and there is no bijection between finite sets of different cardinality.
Problem 1.6.9 Prove that H#% and W % are not isomorphic.
From Problem 1.3.4, we know W% has six elements of order #, whereas H#% has "$ ( =<3 for ! Ÿ 3 "#, and =' ). By Problem 1.6.2, they cannot be isomorphic. Alternatively, notice by Problem 1.3.15 that the only possible orders of elements of W% are in "ß #ß $ß %ß lcm #ß # œ "ß #ß $ß % . However, H#% has an element of order "# (namely <). Proof .
e
kk kk
a bf e
f
Problem 1.6.10 Fill in the details of the proof that the symmetric groups W? and W H are isomorphic if ? œ H as follows: let ) À ? Ä H be a bijection. Define
and prove the following
ab
: À W? Ä WH by : 5 œ ) ‰ 5 ‰ ) " for all 5 − W?
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
(a) : is well-defined, that is, if 5 is a permutation of ? then ) ‰ 5 ‰ ) " is a permutation of H. (b) : is a bijection from W? onto W H. [Find a 2-sided inverse for :.]
a b ab ab
(c) : is a homomorphism, that is, : 5 ‰ 7 œ : 5 ‰ : 7 . Note the similarity to the change of basis or similarity transformations for matrices. Proof .
(a) Notice ) À ? Ä H, 5 À ? Ä ?, and ) " À H Ä ? , so that
) ‰ 5 ‰ ) " À H Ä H. This is a definition of ) ‰ 5 ‰ )" being a permutation of H, so : is indeed well-defined. (b) A two-sided inverse is :" œ ) ‰ 5 " ‰ ) ":
: ‰ :" œ ) ‰ 5 ‰ )" ‰ ) ‰ 5 " ‰ ) " œ ) ‰ 5 ‰ 5 " ‰ ) " œ ) ‰ ) " œ ", and :" ‰ : œ ) ‰ 5 " ‰ ) " ‰ ) ‰ 5 ‰ ) " œ ) ‰ 5 " ‰ 5 ‰ ) " œ ) ‰ ) " œ ".
a b abab a b a b a b a b a b a b a b a b a b a b a b a b a b a ba b a b a b aa ba bb a b a b a ba b a b a b : 5 ‰ 7 œ ) ‰ 5 ‰ 7 ‰ ) " œ ) ‰ 5 ‰ ) " ‰ ) ‰ 7 ‰ ) " œ : 5 : 7 .
(c)
Problem 1.6.11 Let E and F be groups. Prove that E ‚ F z F ‚ E .
: Let +ß , È ,ß + . If : +ß , œ : -ß . (with +ß - − E and ,ß . − F ) then : +ß , œ ,ß + and : -ß . œ .ß - so that ,ß + œ .ß - . But then , œ . and + œ -, so that +ß , œ -ß . . Assume ,ß + − F ‚ E. Then : +ß , œ ,ß + so that :" exists for all elements in F ‚ E. Hence, : is a bijection. Finally, if +ß , ß - ß . − E ‚ F , then (leaving group operations implicit) Proof .
: +ß , -ß .
œ : +-ß ,. œ ,.ß +- œ ,ß + .ß - œ : +ß , : -ß . .
Problem 1.6.12 Let E, F , and G be groups and let K œ E ‚ F and L œ F ‚ G . Prove that K ‚ G is isomorphic to E ‚ L .
a b a b a b a b a a b b a a b b a a b b a a b b a a b b a a b b a b a b a a b b a a b b a a bb aa b b a a bb aaa b baa b bb aa ba b b aa b b a a bb a a ba bb a a bba a bb aa b b aa b b ab ab a b a b a b a b a b aa bb aa bb a b a b a ab b a b a b a b a b ab a b a b a b e a b f ab a b ab ab Proof . In other words, prove E ‚ F ‚ G z E ‚ F ‚ G . Let : À E ‚ F ‚ G Ä E ‚ F ‚ G be defined by : +ß , ß - œ +ß ,ß - . If : +ß , ß - œ : .ß / ß 0 , then +ß ,ß - œ .ß /ß 0 so that + œ . and ,ß - œ /ß 0 and hence , œ / and - œ 0 . But then +ß , ß - œ .ß / ß 0 . Now take +ß ,ß - − E ‚ L . Then : +ß , ß - œ +ß ,ß - so that :" exists for all elements in E ‚ L . Finally, for +ß . − Eß ,ß / − Fß and -ß 0 − G (leaving group operations implicit)
:
+ß , ß -
.ß / ß 0
œ : +ß , .ß / ß -0 œ : +.ß ,/ ß -0 œ +.ß ,/ß -0 œ +.ß ,ß - /ß 0
œ +ß ,ß -
.ß /ß 0
œ : +ß , ß - : .ß / ß 0 .
Problem 1.6.13 Let K and L be groups and let : À K Ä L be a homomorphism. Prove that the image of : , : K , is a subgroup of L . Prove that if : is injective then K z : K . Proof . Let Bß C − : K © L with +ß , − K such that : + œ B and : , œ C. Then : + : , œ : +, , and since +, − K, : +, − : K . Similarly, : , : + − : K . Hence, the operation is closed in " : K . Furthermore, for B − : K , let C − K be such that : C œ B. Then : C " œ : C œ B " − : K , so that : K is closed under inverses. Since L is a group and : K © L, associativity holds. Hence : K is a subgroup of L . If : is injective, then since : K œ 2 − L l b1 − K s.t. : 1 œ 2 , by definition if 2 − : K , then there is 1 − K such that : 1 œ 2. Hence, : is injective. This gives a bijective homomorphism from K to : K , so that K z : K .
Robert Krzyzanowski
e
Solutions to Dummit and Foote
Chapter 1
ab f
Problem 1.6.14 Let K and L be groups and let : À K Ä L be a homomorphism. Define the kernel of : to be 1 − K l : 1 œ "L . Prove that the kernel of : is a subgroup of K . Prove that : is injective if and only if the kernel of : is the identity subgroup of K .
a b a ab b a ab b a b a b
a b a b
Proof . If Bß C − ker :, then : BC œ : B : C œ " L " L œ " L so that BC − ker :. Hence it is closed under the operation. Next, notice "L œ : " K œ : BB" œ : B : B" œ " L : B " œ : B " so that B" − ker :. Hence it is also closed under inverses. Since "K − ker : and associativity follows from the fact K is a group, ker : is itself a group and hence a subgroup of K. If : is injective, then there can only be one element mapped to "L . This must necessarily be "K, so that ker : œ " Ÿ K. Conversely, if ker : œ ", then if " : 1 œ : 2 , 1 Á 2 would give a contradiction, as then "L œ : 1 : 2 œ : 12" so that 12" Á "K is in ker :.
ab ab
ab ab a b
aa bb
Problem 1.6.15 Define a map 1 À ‘# Ä ‘ by 1 Bß C kernel of 1. Proof .
We easily see that
aa b a bb aa aa bb
1 Bß C Bw ß Cw As for the kernel,
1 Bß C
œ 1 B B wß C C w
œ B. Prove that 1 is a homomorphism and find the
bb
aa bb aa bb ea b f aa bb
œ B B w œ 1 Bß C 1 B wß C w .
œ ! Í B œ ! Í ker 1 œ
aa bb
!ß C l C − ‘ .
Problem 1.6.16 Let E and F be groups and let K be their direct product, E ‚ F . Prove that the maps 1" À K Ä E and 1# À K Ä F defined by 1" +ß , œ + and 1# +ß , œ , are homomorphisms and find their kernels. Proof .
We easily see that
aa ba bb aa bb aa bb aa bb aa ba bb aa bb aa bb aa bb aa bb ea b f aa bb ea b f # a b í 1" +ß , +w ß , w
œ 1" ++ wß ,, w
1# +ß , +w ß , w
œ 1# ++ wß ,, w
œ ++w œ 1" +ß , 1" + wß , w , and œ ,, w œ 1# +ß , 1# + wß , w .
Additionally, it's clear that
1" +ß ,
1# +ß ,
In general, 15 À
3−M E3
œ ! Í + œ ! Í ker 1" œ
œ ! Í , œ ! Í ker 1# œ
!ß , l , − F , and +ß ! l + − F .
Ä E5 with +" ß ÞÞÞ È + 5 (for E 3 groups) is a homomorphism with kernel ÖÐ!ß !ß ÞÞÞß +5 ß ÞÞÞß !Ñ l + 5 − E 5× . 5 th term
Problem 1.6.17 Let K be any group. Prove that the map from K to itself defined by 1 È 1" is a homomorphism if and only if K is abelian.
ab ab
ab ab
"
Proof . Call the map :. Then for +ß , − K, : +, œ +, œ , " + ". This equals + ", " œ : + : , if and only if ," +" œ + ", " if and only if ,+ œ +, (multiply by +, on the right and ,+ on the left).
Problem 1.6.18 Let K be any group. Prove that the map from K to itself defined by 1 È 1# is a homomorphism if and only if K is abelian.
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
ab ab f a b a b a b aˆ b a‰ b
ab ab
#
Proof . Call the map :. Then for +ß , − K, : +, œ +, œ +,+,. This equals + #, # œ : + : , if and only if +,+, œ +# , # if and only if ,+ œ +, (cancel the +'s on the left and ,'s on the right).
e
Problem 1.6.19 Let K œ D − ‚ l D 8 œ " for some 8 − ™ . Prove that for any fixed integer 5 " the map from K to itself defined by D È D 5 is a surjective homomorphism but not an isomorphism. 5
œ D"5 D#5 œ : D" : D# . To see it is surjective, 85 5Î5 let D − ‚. By definition, D 8 œ " for some 8 − ™ so thus " œ D 8 so that D "Î5 − K with œ D "Î5 : D "Î5 œ D . The map is not isomorphic because if D Á " is a 5-th root of unity (from complex analysis we know such a number exists for 5 "), then D 5 œ " and "5 œ ". Proof .
Call the map :. First, notice : D" D# œ D" D#
ˆ ‰
ab
aa bb
Problem 1.6.20 Let K be a group and let Aut K be the set of all isomorphisms from K onto K . Prove that Aut K is a group under function composition (called the automorphism group of K and the elements of Aut K are called automorphisms of K).
a b ab ab a ba b a a bb a a b a bb a a bb a a bb a ba b a ba b a ba b a ba b a a b b a a b b ab ab a b ab a b a a b b a ba b ab a ba b a a bb a b a ba b a a bb a b a b a a b ba b a ba b ab aa b ba b a ba a bb a a a bbb aa ba bb a a bba b ab
Sketch. Let :ß < − Aut K . Then : ‰ < À K Ä K , and since composition of bijective functions is bijective, : ‰ < − Aut K so that the set is closed. Furthermore, " ‰ : œ : ‰ " œ :, and the inverse function :" is an isomorphism with the property :" ‰ : œ : ‰ : " œ ". Finally, associativity follows from associativity of functional composition. Proof .
Let :ß < − Aut K . Then : ‰ < À K Ä K . Furthermore, for Bß C − K,
: ‰ < BC œ : < BC
œ: < B < C
œ : < B †: < C
œ : ‰< B † : ‰< C ,
so that : ‰ < is a homomorphism. Assume : ‰ < B œ : ‰ < C , that is, : < B œ : < C . Injectivity of : implies < B œ < C , and injectivity of < implies B œ C. Therefore, : ‰ < is injective. Furthermore, if D − K, then by surjectivity of :, there is a C − K such that : C œ D. By surjectivity of < , there is an B − K such that < B œ C, so that D œ : C œ : < B œ : ‰ < B . Hence, : ‰ < is surjective, and so indeed an isomorphism and thus in Aut K . This proves closure. To show there is an identity, consider " À K Ä K given by 1 È 1. Then for B − K,
:‰" B œ: " B
œ : B and " ‰ : B œ " : B œ : B .
By definition " is the identity in Aut K . Since : is a bijection, it has an inverse function :" such that : ‰ :" B œ :" ‰ : B œ B for all B − K . This is the definition of : " being the inverse of : in Aut K . Finally, composition of functions is associative in general, and for :ß <ß 9 − Aut K with B − K,
:‰< ‰9 B œ :‰< 9 B
This means that Aut K composition.
œ: < 9 B
œ : < ‰9 B
œ 9 ‰ < ‰9
B .
satisfies all the group axioms so that it is indeed a group under function
Problem 1.6.21 Prove that for each fixed nonzero 5 − the map from to itself defined by ; È 5; is an automorphism of .
a b a b a b a b
a b a b
ab ab
Proof . Call the map :. For :ß ; − , : : ; œ 5 : ; œ 5: 5; œ : : : ; . To show bijectivity, notice : : œ : ; means 5: œ 5; so that division gives : œ ;; if ; − , then ;Î5 − gives : ;Î5 œ 5 ;Î5 œ ; so that : is surjective.
Problem 1.6.22 Let E be an abelian group and fix some 5 − ™. Prove that the map + È +5 is a homomorphism from E to itself. If 5 œ ", prove that this homomorphism is an isomorphism.
Robert Krzyzanowski
+"
Solutions to Dummit and Foote
ab
Chapter 1
5
Proof . If +ß , − E, then +, œ + 5 , 5 since E is abelian (Problem 1.1.24). If 5 œ ", then if œ , " , then " œ ++" œ +, " œ ,, " so that cancellation on the right by , " gives + œ ,. To show
a b
surjectivity, notice if + − E, then of course +"
"
œ +.
ab
Problem 1.6.23 Let K be a finite group which possesses an automorphism 5 such that 5 1 œ 1 if and only if 1 œ ". If 5 # is the identity map from K to K, prove that K is abelian (such an automorphism 5 is called fixed point free of order #).
a b ab ab abab a b ab ab a b a a bb a b a a bb
ab
Proof . We claim there is a bijection B È B" 5 B . To show this map is injective, let B " 5 B œ C" 5 C . Then 5 B œ BC " 5 C so that 5 B 5 C " œ 5 BC " œ BC ". Since 5 is fixed point free, BC" œ ", so that C œ B. Since this is a finite map from K Ä K and it is injective, it must be surjective. In other words, each element 1 − K can be written 1 œ B" 5 B for some B − K.
ab
To show that K is abelian, notice if 1 œ B" 5 B for any 1 − K, then
5 1 œ 5 B" 5 B
œ5 B
"
B œ B" 5 B
"
œ 1 " ,
and apply Problem 16.17 (which states that a group possessing the homomorphism 1 È 1" is abelian).
kk
Problem 1.6.24 Let K be a finite group and let B and C be distinct elements of order # in K that generate K . Prove that K z H#8 , where 8 œ BC . Proof .
Let > œ BC . Then from Problem 1.2.6, these elements give a presentation
Ø > ß B l >8 œ B# œ " , BC œ >B œ B> " Ù .
a b a b a b
Notice this is the standard presentation of H#8. This is enough to guarantee isomorphism, but we can show this explicitly. Let : À H#8 Ä K given by : =3< 4 œ B 3> 4. Then B and > in K satisfy all the relations that = and < do in H#8. Hence, if : =3 <4 œ : = 8< 7 with 3ß 8 − !ß " ß 4ß 7 − !ß ÞÞÞß 8 " , then B 3> 4 œ B 8> 7. Assume 3 Á 8. Without loss of generality, let 3 œ ! and 8 œ ". Then >4 œ B> 7, so that B œ > 47, which is a contradiction, since B and > satisfy the same relations in K and = and < do in H#8, specifically = Á < 3 for any 3 (respectively, B Á >3 for any 3, e.g., 3 œ 4 7). Hence, 3 œ 8. But then either way (by multipling B on the Î4 left if there is no B), this means B>4 œ B>7. However, again, we know H#8 has the relation =< 3 Á =< 4 for 3 ´ 4 7 Î 7 mod 8 . Since 4ß 7 − !ß ÞÞÞß 8 " , we have 4 œ 7. Hence, mod 8 , which induces B> Á B> for 4 ´ 3 4 8 7 3 œ 8 and 4 œ 7, so that = < œ = < . This shows injectivity. Surjectivity is trivial, since if B 3> 4 − K (and we know every element can be written this way since K has the same relations as H#8), then : =3 < 4 œ B 3> 4. Finally, homomorphism follows from:
a b a
e f
a b
b a
b a
e
e
f
f
b
a b
a ba b
: =3 <4 =8 <7 œ : =3 =8 <4< 7 œ : =38 <74 œ B38 >74 œ B3 B8 >4> 7 œ B3 >4B 8 >7 œ : =3 <4 : =8 <7 ,
where the penultimate equality follows again from the fact B and > obey the same relations in K as = and < do in H#8 . Problem 1.6.25 Let 8 − ™ , let < and = be the usual generators of H#8 and let ) œ #1Î8.
Œ
ab
cos ) sin ) is the matrix of the linear transformation which rotates the sin ) cos ) B , C plane about the origin in a counterclockwise direction by ) radians.
(a) Prove that the matrix
(b) Prove that the map : À H#8 Ä KP# ‘ defined on generators by
ab Œ
: < œ
cos ) sin )
sin ) cos )
ab Œ
and : = œ
! " " !
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
ab
extends to a homomorphism of H#8 into KP# ‘ .
(c) Prove that the homomorphism : in part (b) is injective.
a b
a
b
Proof . (a) Let Bß C − ‘# . Using polar coordinates, this point is < cos <ß < sin < . Rotating by ) means the new point is (represented as a vector)
Πaa bb Πaa < cos < ) < sin < )
bb Œ
< cos < cos ) sin < sin ) < cos < sin ) sin < cos )
œ
œ
B cos ) C sin ) B sin ) C cos )
Œ œ
sin )
cos ) sin )
cos )
Œ
B . C
(b) Notice that from the previous exercise, the matrix < is mapped to rotates the Bß C plane by ) radians, so that applying it : times leads to a rotation by : ) radians, that is,
Œ
sin )
cos ) sin )
cos )
Π:
œ
cos :) sin :)
sin :) cos : )
ab a b ˆ ab ‰ Œ Œ Œ Œ Œ Œ Œ Œ Œ Œ a ba b ab Œ Œ Œ Œ Œ Œ Œ Œ Œ Œ Œ Œ Œ
for all : − ™ . This can be proved using a simple inductive argument, but we will take our intuition as evidence. Then if we let = 4 œ 4 if 8 œ ! and 4 if 8 œ ", 3 4 8 7
: =<= <
œ: =
38 = 4 7
<
œ
! " " !
œ
! " " !
3
3
38
! " " !
œ
! " " !
8
cos ) sin )
sin )
cos ) =Ð4Ñ
sin )
cos ) sin )
cos )
=Ð4Ñ
cos )
=Ð4Ñ7
sin )
cos ) sin )
! " " !
8
cos ) sin )
sin )
cos ) sin )
sin )
7
cos )
7
cos )
œ : =3 < 4 : = 8 7 ,
where the middle step is trivial if 8 œ ! (so that = 4 œ 4), and for 8 œ ",
! " " !
8
sin )
cos ) sin )
œ
! " " !
œ
sin 4) cos 4)
=Ð4Ñ
œ
cos )
cos -4) sin -4) cos 4) sin 4)
! " " !
sin -4) cos -4) œ
cos 4) sin 4)
cos )
! " " !
œ
sin 4) cos 4)
4
sin )
cos ) sin )
cos 4) sin 4)
! " " !
œ
sin 4) cos 4)
cos ) sin )
=Ð4Ñ
sin ) cos )
! " " !
8
.
(c) Assume
a b a b ΠΠΠΠe f e f ΠΠΠΠ3 4
: =<
8 7
œ: = <
, or
! " " !
3
sin )
cos ) sin )
cos )
4
œ
! " " !
8
cos ) sin )
sin )
7
cos )
where 3ß 8 − !ß " and 4ß 7 − !ß ÞÞÞß 8 " . However, from the initial remark in the solution of part (b),
! " " !
Furthermormore, notice
3
cos 4) sin 4)
sin 4) cos 4)
œ
! " " !
8
cos 7) sin 7)
sin 7) Þ cos 7)
(1)
Robert Krzyzanowski
Solutions to Dummit and Foote
ΠΠ! " " !
cos :) sin :)
sin :) cos :)
Œ œ
sin :) cos :)
Chapter 1
cos : ) . sin : )
Hence, if 3 Á 8 and assuming without loss of generality that 3 œ " so that 8 œ !, we would need
ÚÝ ÛÝ Ü
cos 7) œ sin 4) cos 4) œ sin 7) cos 4) œ sin 7) cos 7) œ sin 4)
(2)
e e f
f˜ ™
in order for the two matrices in (1) to be equal. However, 4ß 7 − !ß ÞÞÞß 8 " , so ! Ÿ 4) # 1 and ! Ÿ 7) #1. Notice the first and last equation in (2) force 4) − !ß 1 so 7) − 1# ß $#1 , and the second and third equation give 7) − !ß 1 and so 4) − 1# ß $#1 . Clearly, these four equations have no solution, so that 3 œ 8. We may multiply by the inverse of !" "! if 3 œ 8 œ ". Then
˜ ™ ˆ‰ Œ
e f Œ
cos 4) sin 4)
sin 4) cos 4)
cos 7) sin 7)
œ
sin 7) cos 7)
a
b
so that we have the equations cos 4) œ cos 7) and sin 4) œ sin 7). This means 4) ´ 7 ) mod # 1 , but since 0 Ÿ 4) #1 and ! Ÿ 7) #1, this means 4) œ 7) and so 4 œ 7. † Hence, 3 œ 8 and 4 œ 7 so that =3 <4 œ = 8< 7 and the map : is injective.
ab
Problem 1.6.26 Let 3 and 4 be generators of U) described as in Section 5. Prove that the map : from U) to KP# ‚ defined on generators by
a b È È a b Œ -"
: 3 œ
!
!
and : 4 œ
-"
! " " !
extends to a homomorphism. Prove that : is injective. Proof .
First, we need to know the values of the ::
a b a b ab ab Œ Œ ab a b a b a b a b È È Œ È È ababa b a b a b a b a b a b a b a b a b e f a b ab ab a b a b È È È È Œ È È a b a b #
: " œ : 4
! " " !
œ:4:4 œ
-"
: 5 œ : 34 œ : 3 : 4 œ
#
" ! ! "
œ
!
!
! " " !
-"
œ " † "KP# ‚ , and
œ
!
-"
-"
!
,
where : 3 ß : 4 ß : 5 can easily be seen to be : 3 ß : 4 ß and : 5 seen as multiplying the scalar " − ‚ by each matrix (the action of : " ). Then : B † „ " œ : B : „ " œ : „ " † B for each B − 3ß 4ß 5 . We already know : 34 œ : 3 : 4 . For all the others, : 43 œ : 5 œ
†There
!
-"
-"
!
œ
-"
!
-"
œ
!
a
b
! " " !
-"
!
!
-"
œ: 4 : 3 ,
also more fun ways of showing this. For example, if we let 4) ß 7) − Ò1ß 1 Ñ for a moment, then the injectivity of arcsin À Ò"ß "Ñ Ä Ò1 ß 1 Ñ gives 4) œ 7). Then take 4) and 7) mod #1 , so that 4 œ 7 . Alternatively, we could have noticed . ) œ ..) sin 7) so that 4 cos 4) œ 7 cos 7 ), but since cos 4) œ cos 7)ß 4 cos 4 ) œ 7 cos 4 ) so that 4 œ 7 . .) sin 4
a b a b
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
a b a b Œ È È È È a b a b Œ Œ È È È a b a b È È Œ È È a b a b a b È È È È Œ : 53 œ : 4 œ
: 35 œ : 4 œ
! " " !
! " " !
: 45 œ : 3 œ
œ
-"
!
: 54 œ : 3 œ : 3 œ
eab
!
œ
-"
! " " !
œ
!
-"
!
!
f
-"
-"
!
!
œ
-"
!
!
-"
-"
-" !
!
!
!
!
-"
! " " !
œ
-"
-"
-"
-"
!
a b ab È a b a b
œ: 5 : 3 ,
-" -" !
œ: 3 : 5 ,
ab a b a b ab
œ : 4 : 5 , and
! " " !
œ: 5 : 4 .
ab ab
We can look at the matrices : B À B − U) œ im : and see that there are no two same : B œ : C for : so that is injective by inspection. B Á C,
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Chapter 1.7 Problem 1.7.1 Let J be a field. Show that the multiplicative group of nonzero elements of J (denoted by J ‚ ) acts on the set J by 1 † + œ 1+, where 1 − J ‚ , + − J and 1+ is the usual product in J of the two field elements (state clearly which axioms in the definition of a field are used ). Proof . Let 1" ß 1# − J ‚ and + − J (as a set). Then 1# † + œ 1 #+ − J where 1 #+ is the product in J as a field. Notice this is possible since multiplication in J as an abelian group is only defined for non-zero elements, and 1# − J ‚ implies 1# Á !. Furthermore, then 1" † 1# + œ 1" 1#+ , where 1#+ is taken as in J and 1" is taken as in the group J ‚. Since multiplication in the field J is a group, and similarly 1" Á !, we can say 1" 1# + œ 1" 1# + by group associativity. Hence, 1" † 1# † + œ 1 "1 # † +. Finally, if + − J , then for ‚ ‚ the " − J , " † + œ " + œ +, since " is the multiplicative identity in the field J. This shows J satisfies definition of being an action on J .
a b a b a b a b
a b aa b b
Problem 1.7.2 Show that the additive group ™ acts on itself by D † + œ D + for all Dß + − ™.
a b a b a b a b
Proof . Let D" ß D# − ™ as a group and + − ™ as a set. Then D# † + œ D# + where the sum is computed as the group operation in ™ß . We then get
a b a b
a b
D" † D# + œ D" D# + œ D " D # + œ D " D # † +
since addition in ™ as a group is associative. Notice since the group operation is addition here, we showed it explicitly when writing D" D# † +. Finally, "™ † + œ ! + œ +. Note. These first two problems were trivial, but showed the care that must be taken when considering the operations being applied (e.g., even though D + was an element of the set ™ , the actual element was the group operation of ™ß applied to the elements Dß + − ™ß ). From now on, we will drop such pedantry and immediately become less formal by treating the operations implicitly (unless some extreme trickery is at hand!).
a b
a b a b b a a b b a ba b
Problem 1.7.3 Show that the additive group ‘ acts on the Bß C plane ‘ ‚ ‘ ,C < † Bß C œ B
a b a a bb a b aa b b a b ba b a
Proof .
Let <ß = − ‘ and Bß C − ‘ ‚ ‘. Then
< † = † Bß C
a
œ < † B =Cß C œ
B =C
and " ‘ß † Bß C œ B !Cß C œ Bß C .
Problem 1.7.4 Let K be a group acting on a set E and fix some + − EÞ Show that the following sets are subgroups of K .
(a) the kernel of the action, (b)
e
f
1 − K l 1+ œ + this subgroup is called the stabilizer of + in K .
a b a ab ab b
Proof . (a) Let 1ß 1" ß 1# be in the kernel of the action. Then 1 † + œ 1" † + œ 1 # † + œ + for all + − E. First, notice that 1" 1# is also in the kernel of the action, since 1" 1# † + œ 1" † 1 # † + œ 1 " † + œ +. Thus, the kernel is closed under the group operation. Additionally, 1" † + œ 1" † 1 † + œ 1 "1 † + œ " † + œ +, so that the kernel is closed under inverses. Hence, it is a subgroup of K.
e
f
(b) Replace "kernel [of the action]" by this set and remove "for all + − E" from the proof of part (a) and this shows 1 − K l 1+ œ + is a subgroup of K.
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
Problem 1.7.5 Prove that the kernel of an action of the group K on the set E is the same as the kernel of the corresponding permutation representation K Ä W E . Proof . Let 1 be in the kernel of the action. Then 1+ œ + a+ − E so the permutation representation of 1 is the trivial permutation " − WE. That is, if we let : À K Ä WE be the map that takes elements in K to their permutation representations, : 1 œ "W E. But this is the definition of 1 being in the kernel of the permutation representation (see Problem 1.6.14). Conversely, let 1 − K such that : 1 œ "W E. Then : 1 + œ +. However, by definition : 1 + œ 1 † +, which means 1 is in the kernel of the action.
ab a ba b
ab
a ba b
Problem 1.7.6 Prove that a group K acts faithfully on a set E if and only if the kernel of the action is the set consisting only of the identity. Proof . Assume K acts faithfully on E, that is, permutation representations are unique. Notice that for " − K (the identity), by definition " † + œ + a+ − E. Since K acts faithfully, this must be the only element in K such that 1 † + œ +. But this means the kernel of the action is " . Now assume the converse. By way of contradiction, assume 1" + œ 1# + œ +w for all + − E for different 1 " and 1 # in K. But then
ef
a b a b ab e f
a b a b ab f e f
w " " " w " " 1" " + œ 1" 1" + œ 1 " 1" + œ " + œ + and 1 # + œ 1 # 1 #+ œ 1 # 1 # + œ " + œ + . w " w That is, 1" " + œ 1# + œ +. However, since 1" + À + − E œ 1 #+ À + − E is a permutation on E, this w " w means that +w œ 1" + œ 1# + À + − E is also a (indeed the same) permutation. Hence, 1 " " + œ 1 # + œ + for " " " " all +w − E. However, if 1" Á 1# , then 1" " Á 1# so that one of 1" and 1# is not the identity, say 1" . But w w then 1" " + œ + for all + − E gives a contradiction, since we assumed the kernel of the action is " , and " 1" " Á " so that 1" is not in the kernel.
e
ef
Problem 1.7.7 Prove that in Example 2 in this section the action is faithful. Proof .
For completeness, Example 2 is copied here below:
Example 2. The axioms for a vector space Z over a field J include the two axioms that the multiplicative group J ‚ act on the set Z . Thus vector spaces are familiar examples of actions of multiplicative groups of fields where there is even more structure ( in particular, Z must be an abelian group) which can be exploited . In the special case when Z œ ‘8 and J œ ‘ the action is specified by
a
a
b a
! <" ß ÞÞÞß < 8 œ !< "ß ÞÞÞß !< 8
b
b
for all ! − ‘, <" ß ÞÞÞß < 8 − ‘8 , where !< 3 is just multiplication of two real numbers. è
Assume that the action is not faithful. That is, there are !" ß !# − ‘ so that
a
b
a
b a b a b a b a b a b Š ‹ a b ˆ ‰ e f Š ‹ k k e f e a b a bf
!" <" ß ÞÞÞß <8 œ !# < "ß ÞÞÞß <8 , or ! "< "ß ÞÞÞß ! "< " œ ! #< "ß ÞÞÞß ! #< 8
for all <" ß ÞÞÞß < 8 − ‘8 . Pick some < "ß ÞÞÞß < 8 Á ! with say < 5 Á ! for some " Ÿ 5 Ÿ 8. Then comparing the 5-th components, !" <5 œ !#<5 , so that if we take <"5 − ‘, then using the definition of an action, " <5
!" <5 œ
" <5 ! "
<5 œ
" <5
! #< 5 œ
" 5 !# < 5
Since the elements here are computed as multiplication of two real numbers, that !" œ !# . By definition, this means the action must be faithful.
.
" <5 ! 3
<5 œ !3 for 3 − "ß # so
Problem 1.7.8 Let E be a nonempty set and let 5 be a positive integer with 5 Ÿ E . The symmetric group WE acts on the set F consisting of all subsets of E of cardinality 5 by 5 † +"ß ÞÞÞß + 5 œ 5 + " ß ÞÞÞß 5 + 5 .
(a) Prove that this is a group action.
Robert Krzyzanowski
Solutions to Dummit and Foote
Chapter 1
ab a b e f e f a e fb e a b a bf e a a bb a a bbf ea ba b a ba bf a b e f e f e f e a b a bf e f aa bb ee ff ee ff aa bb ee ff ee ff aa bb ee ff ee ff aa bb ee ff ee ff aa bb ee ff ee ff aa bb ee ff ee ff e ab f a b a b e f e f a b a a bb a a b a bb Š a a bb a a bb‹ Ša ba b a ba b‹ a b a b a b a b a a b a bb a b
(b) Describe explicitly how the elements " # and " # $ act on the six #-element subsets of "ß #ß $ß % . Proof .
(a) Let 5" ß 5# − WE . Then for all +" ß ÞÞÞß + 5 − F ,
5" † 5# † + "ß ÞÞÞß + 5
œ 5" † 5# + " ß ÞÞÞß 5# + 5 œ
œ 5 " 5# + " ß ÞÞÞß 5 " 5 # + 5
5" 5# +" ß ÞÞÞß 5" 5# + 5
œ 5"5# † + "ß ÞÞÞß + 5 .
To show the second property of group actions, take any +" ß ÞÞÞß + 5 − F; then
" W E † +"ß ÞÞÞß + 5 œ " WE + " ß ÞÞÞß " WE + 5
(b)
œ + "ß ÞÞÞß + 5 .
" # † "ß # œ #ß " , " # † "ß $ œ #ß $ ß " # † "ß % œ #ß % ß " # † #ß $ œ "ß $ ß " # † #ß % œ "ß % ß " # † $ß % œ $ß % .
" # $ † "ß # œ #ß $ , " # $ † "ß $ œ #ß " ß " # $ † "ß % œ #ß % ß " # $ † #ß $ œ $ß " ß " # $ † #ß % œ $ß % ß " # $ † $ß % œ "ß % . Notice that in general, the stabilizer (see Problem 1.7.4) of a subgroup F © E acted upon by 5 − WE is given by 5 − WE À 5 , œ , a, − F , that is, the permutations that fix all , − F. Problem 1.7.9 Do both parts of the preceding exercise with "ordered 5 -tuples" in place of "5 -element subsets," where the action on 5 -tupes is defined as above but with set braces replaced by parentheses ( note that, for example, the #-tuple "ß # and #ß " are different even though the sets "ß # and #ß " are the same, so the sets being acted upon are different ). Proof .
(a) Let 5" ß 5# − WE . Then for all +" ß ÞÞÞß + 5 − F,
5" † 5# † +"ß ÞÞÞß + 5
œ 5" † 5# + " ß ÞÞÞß 5# + 5 œ
œ 5 " 5# + " ß ÞÞÞß 5 " 5 # + 5
5" 5# +" ß ÞÞÞß 5" 5# + 5
œ 5"5# † + "ß ÞÞÞß + 5 .
To show the second property of group actions, take any +" ß Þ ÞÞß + 5 − F; then
" W E † + "ß ÞÞÞß + 5 œ " WE + " ß ÞÞÞß " WE + 5
(b)
œ + "ß ÞÞÞß + 5 .(b)
aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa b b aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bb aa bbaa bb aa b b a a b b " # † "ß # œ #ß " , " # † #ß " œ "ß # , " # † "ß $ œ #ß $ ß " # † $ß " œ $ß # ß " # † "ß % œ #ß % ß " # † %ß " œ %ß # ß " # † #ß $ œ "ß $ ß " # † $ß # œ $ß " ß " # † #ß % œ "ß % ß " # † %ß # œ %ß " ß " # † $ß % œ $ß % ß " # † %ß $ œ %ß $ ß " # † "ß " œ #ß # ß " # † #ß # œ "ß " ß " # † $ß $ œ $ß $ ß " # † %ß % œ %ß % Þ
" # $ † "ß # " # $ † "ß % " # $ † #ß % " # $ † "ß "
e
œ œ œ œ
#ß $ , #ß % ß $ß % ß #ß # ß
"# $ † "#$ † "#$ † " # 3 †
ab
#ß " %ß " %ß # #ß #
œ œ œ œ
$ß # , " # $ %ß # ß " # $ %ß $ ß " # $ $ß $ ß " # $
† † † †
"ß $ œ #ß " ß " # $ † $ß " œ "ß # ß #ß $ œ $ß " ß " # $ † $ß # œ "ß $ ß $ß % œ "ß % ß " # $ † %ß $ œ %ß " ß $ß $ œ "ß " ß " # $ † %ß % œ %ß % Þ
f
a
b
Notice that in general, the stabilizer (see Problem 1.7.4) of a sub-tuple F œ ," ß ÞÞÞß , 5 © E acted upon by 5 − WE is given by 5 − WE À 5 ,3 œ ,3 a" Ÿ 3 Ÿ 5 , that is, the permutations that fix all the components of the tuple, ,3 . Problem 1.7.10 With reference to the preceding two exercises determine:
(a) for which values of 5 the action of W8 on 5-element subsets is faithful, and (b) for which values of 5 the action of W8 on ordered 5 -tuples is faithful.
Robert Krzyzanowski Proof .
Solutions to Dummit and Foote
Chapter 1
Call the set we will be taking subsets of E. We will make use of the following lemma:
e f
ab
Lemma. Let E Á g and 5 À E Ä E be a permutation. If 5 also permutes finite subsets F © E and G © E (that is, for each W − Fß G ß B − W , we have 5 B − W ), then 5 permutes F G.
ab
ab ab
Let 5 be a permutation of E, F and G . Assume there is an B − F G such that 5 B Â F G, say 5 B − F without loss of generality. This contradicts the fact 5 permutes G (since that fact says 5 B − G). Hence, 5 permutes F G . û Proof .
+
If E Á gß 5 À E Ä E is a permutation, and E 3 © E are subsets of E, then, 5 is also a Corollary. permutation on E3. û
ef
Since E is finite, index it by E œ +3
8 3œ"
kk ab
where 8 œ E .
(a) If 5 œ !, every permutation of g is g, so the action is not faithful. From Problem 1.7.6, all we need to determine is the kernel of the action. If 5 œ ", then if " † W œ 5 † W œ W for each one-element subset W of E, then 5 +3 œ +3 for each " Ÿ 3 Ÿ 8, so that 5 + 3 œ + 3 for + 3 − E; then by definition 5 œ ". Thus, for 5 œ " the action is faithful. Let " 5 8. Assume there is a non-trivial permutation 5 on E so that for each W − W © E À W œ 5 it is true " † W œ 5 † W œ W . For each " Ÿ 3 Ÿ 8 and " Ÿ 4 Ÿ 8 with 3 Á 4, let W34 © E +3 so that + 4 − W 34 with W 34 œ 5 . We know such a set exists for each 3ß 4 with 3 Á 4, since W34 œ 5 Ÿ 8 " œ E + 3 . The fact 5 " guarantees W 34 is non-empty. Then we claim that
a e fb e f e e fk k f k k kk k e fk + f 4 ³
"Ÿ3Ÿ8
ef
W34 œ +4
3Á4
for " Ÿ 4 Ÿ 8.
e f ae f b e f
First, notice +4 − f 4 since by definition +4 − W 34 for each W 34. Assume there is some other + < − W 34. But +<  W<4 and since f4 © W<4, this is a contradiction. Thus, f 4 œ + 4 . Then apply the lemma to the subsets W34Þ This means 5 is a permutation on f4 œ +4 . Hence, 5 + 4 œ + 4 and thus 5 + 4 œ + 4. This holds for each " Ÿ 4 Ÿ 8, so that 5 œ ". We have shown explicitly that " is the only permutation in the kernel of the action, so that for " 5 8 the action is faithful. Finally, if 5 œ 8, the only 8-element subset of E is E itself, and " † E œ 5 † E œ E is true for any permutation 5, since 5 † E œ im 5 œ E, so that the action is not faithful.
ef
ab
In conclusion, if 5 œ ! or 5 œ 8 then the action is not faithful, and otherwise ( " 5 8) it is faithful. This stems from the fact there are at least 8 5-element subsets when the latter condition holds, some of which we can intersect and apply the lemma to. (b) If 5 œ !, every permutation of the !-tuple is the !-tuple, so the action is not faithful. Otherwise, again from Problem 1.7.6, we only need to determine whether the kernel of the action is just the identity. For 5 !, let the 5-tuple X 3 (" Ÿ 3 Ÿ 8) be such that each component of X5 is + 3. In other words, X 5 is just the tuple +3 ß ÞÞÞß +3 where + 3 is written 5 times. Assume the action is not faithful, that is, there is a permutation 5 so that " † X œ 5 † X œ 5 for each 5-tuple X . In particular, this is true for each X 3, so that
a
b
a
b a a b a bb a
5 † +3 ß ÞÞÞß + 3 œ 5 + 3 ß ÞÞÞß 5 + 3
ab
b
œ + 3ß ÞÞÞß + 3 ,
and hence component-wise 5 +3 œ +3. But we can do this for each " Ÿ 3 Ÿ 8, so 5 œ ". Then the kernel of the action is just ", so the action is faithful. Therefore, the action is faithful for all ! 5 Ÿ 8 and not when 5 œ !. Problem 1.7.11 Write out the cycle decomposition of the eight permutations in W % corresponding to the elements of H) given by the action of H) on the vertices of a square (where the vertices of the square are labelled as in Section 2).
Robert Krzyzanowski Proof .
a
Solutions to Dummit and Foote
Chapter 1
Label the vertices as below.
a
ba a b a bab b a b
b a
ba
b a
b
Then the rotations <ß <# ß <$ correspond to the maps "ß #ß $ß % È #ß $ß %ß " ß "ß #ß $ß % È $ ß %ß "ß # ß and "ß #ß $ß % È %ß "ß #ß $ , respectively. Hence, the permutations corresponding to the actions by "ß <ß <# ß and <$ are "ß " # $ % ß " $ # % , and " % $ # , respectively. Similarly, if we let = be the reflection with fixed points in the upper right and lower left corner ( # and % in the original square), we can find the permutations corresponding to the actions by =ß =<ß =< #ß and =< $ to be " $ ß " % # $ ß # % ß and " # $ % , respectively.
a b a ba b a b
a ba b
Problem 1.7.12 Assume 8 is an even positive integer and show that H#8 acts on the set consisting of pairs of opposite vertices of a regular 8-gon. Find the kernel of this action (label vertices as usual).
e
f e
Label the vertices as in Section 2 and let 8 œ #7. Then the pairs partition the set of vertices into 3ß 3 7 , for " Ÿ 3 Ÿ 7 . Let = be the reflection that fixes the points that " and 7 are at originally. Assume =<5 is in the kernel for some " Ÿ 5 8. But, notice that then Proof .
f
e
f
e
f e e f e e f e f e f e f e e ffe f
f
=<5 † "ß 7 " œ < 5 = † "ß 7 " œ < 85 † "ß 7 " œ 8 " 5ß 8 " 7 5 ,
fe f
where we let 8 "ß 8 #ß ÞÞÞß #8 be the same points as "ß #ß ÞÞÞ ß 8. However, notice 8 " 5 is not " or 7 " unless 5 œ 8 (i.e., =<5 œ =) or 5 œ 7. In the former case, = † #ß 7 # œ 8 "ß 7 so that = does not fix all vertices either. In the latter case, =<7 † #ß 7 # œ < 7= † #ß 7 # œ < 7 † 8 "ß 7 œ 7 "ß 8 so that =<7 does not fix all vertices either. Hence, no element of the form =<5 can be in the kernel of the action.
e
f
f
Now consider <5 for " Ÿ 5 8 with 5 Á 7. Then < 5 † "ß 7 " œ 5 "ß 5 7 " so that these two sets are not equal unless 5 œ 8 (so that <5 œ ") or 5 œ 7. The former is the trivial case (the identity), so consider the latter. We have <7 † 3ß 3 7 œ 3 7ß 3 #7 œ 3ß 3 7 for each " Ÿ 3 Ÿ 7. Hence, <5 is in the kernel of the action. In conclusion, the kernel is "ß <7 . Notice this also means the action is not faithful.
e
Problem 1.7.13 Find the kernel of the left regular action . Proof . Let K ‚ K Ä K be given by 1 † + œ 1+ as group multiplication in K (for +ß 1 − K). Then the kernel of this left regular action is the set of 1 − K such that 1+ œ + for all + − K. However, notice that for any + − K , 1+ œ + implies by right cancellation that 1 œ ". Thus, the kernel of the left regular action is just the identity (so that it is a faithful action).
Problem 1.7.14 Let K be a group and let E œ K . Show that if K is non-abelian then the maps defined by 1 † + œ +1 for all 1ß + − K do not satisfy the axioms of a (left) group action of K on itself .
a b a b
a b
Proof . If K is non-abelian, it has two non-commuting elements 1" ß 1#. If the given map was a group action, then if we fix + − K, 1" † 1# † + œ 1 " † +1 # œ +1 #1 " should be equal to 1 "1 # † + œ +1 "1 #. But this is saying +1# 1" œ +1" 1# which by left cancellation implies 1# 1" œ 1" 1#, contradicting the fact 1" and 1# are non-commuting elements.
Problem 1.7.15 Let K be any group and let E œ K . Show that the maps defined by 1 † + œ +1 " for all 1ß + − K do satisfy the axioms of a (left) group action of K on itself .
Robert Krzyzanowski Proof .
Solutions to Dummit and Foote
Let 1" ß 1# − K. Then
a b a b
Chapter 1
a b a b
1" † 1# † + œ 1" † +1" œ +1#"1"" œ + 1 "1 # #
a b
and if we let " be the identity in K, then " † + œ + "" œ +.
"
œ 1 "1 # † + ,
Problem 1.7.16 Let K be any group and let E œ K . Show that the maps defined by 1 † + œ 1+1 " for all 1ß + − K do satisfy the axioms of a (left) group action (this action of G on itself is called conjugation). Proof .
Let 1" ß 1# − K. Then
a b a b
a ba b a b
1" † 1# † + œ 1" † 1#+1" œ 1" 1# +1#"1"" œ 1 "1# + 1 "1 # #
aba b
and if we let " be the identity in K, then " † + œ " + "" œ +.
"
œ 1 "1 # † + ,
Problem 1.7.17 Let K be a group and let K act on itself by left conjugation, so each 1 − K maps K to K by
B È 1B1" .
kk
For fixed 1 − K, prove that conjugation by 1 is an isomorphism from K onto itself (i.e., is an automorphism of K). Deduce that B and 1B1" have the same order for all B in K and that for any subset E of K , E œ 1E1" (here 1E1" œ 1+1" l + − E ).
k k
e
f
a ba b a b
Proof . Let Bß C − K . Then 1BC1 " œ 1B1 " 1C1 " œ 1B1" 1C1 " so that conjugation is a homomorphism. Now assume 1B1" œ 1C1". Then right and left cancellation yield B œ C. Finally, if C − K, then if we conjugate B œ 1" C1 − K we get 1B1" œ 1 1" C1 1" œ C. This shows conjugation is a bijection, so in conclusion it is an isomorphism from K to itself. From Problem 1.6.2, we know B and 1B1" must have the same order for all B in K. Finally, if B − E, then B − 1E1 " by definition. Conversely, if B − 1E1" , then by definition B œ 1C1" for some C − E so that C œ 1 "B1. But since C − E, by definition 1C1" − 1E1" ; but 1C1" œ 1 1" B1 1" œ B. Hence, conjugation gives a bijection from E to 1E1 ". But since conjugation is a homomorphism on K, it must be a homomorphism on E, so that it is an isomorphism from E to 1E1" . But if E z 1E1" , then E œ 1E1 " .
a b
kk k k
Problem 1.7.18 Let L be a group acting on a set E . Prove that the relation µ on E defined by
+µ,
if and only if
+ œ 2,
for some 2 − L
is an equivalence relation. (For each B − E the equivalence class of B under µ is called the orbit of B under the action of L . The orbits under the action of L partition the set E).
ab a b
Proof . To show reflexivity, notice by one of the action axioms that + œ " † +, so that + µ +. Now let +ß , − E and assume + µ ,. Then + œ 2, for some 2 − L . But then 2 " + œ 2 " 2, œ 2 " 2 , œ ", œ ,, so that , µ +. Finally, assume +ß ,ß - − E with + µ , and , µ -. Then + œ 2, for some 2 − L and , œ 1- for some 1 − L . Hence, + œ 2, œ 2 1- œ 21 - so that + µ -.
ab ab
Examples. Since the book provides none, the following examples are given, intended to introduce a more intuitive understanding of the notion of an orbit.
ef
(1) If L œ E œ K for K a group, then for +ß , − E, + œ 2, for some 2 − L if and only if 2 œ ". Therefore, the orbit of each element B for this action is just B . Call this the identity orbit. Then, for example, this implies the action given in Problems 1.7.2 induces the identity orbit on each element in ™. (2) The actions given in Problem 1.7.11 and 1.7.12 induce the orbit of the set itself for each element. In other words, all elements in this action are equivalent under the relation given in the previous problem (since we can "unrotate" and "unreflect" any rotation and reflection of the vertices). Additionally, if the group in
Robert Krzyzanowski
Solutions to Dummit and Foote
e
Chapter 1
f
Problem 1.7.11 was taken to be the subgroup "ß <ß < # ß < $ © H), then the rotations of the square would be in one equivalence class (orbit), and the reflections would be in another (so that the action has two distinct orbits).
a b
a
b
ea
b
f
(3) If L œ ‘‚ ß ‚ and E œ ‘8, then the orbit of + "ß ÞÞÞß + 8 − E is ! + "ß ÞÞÞß + 8 À ! − L . Therefore, we can say if 8 " then EÎ µ z ‘ 8"(real projective space), where is the equivalence relation in the previous problem. µ
Problem 1.7.19 Let L be a subgroup of the finite group K and let L act on K (here E œ K ) by left multiplication. Let B − K and let b be the orbit of B under the action of L . Prove that the map
LÄb
defined by
kk
2 È 2B
is a bijection (hence all orbits have cardinality L ). From this and the preceding exercise deduce Lagrange's Theorem:
kk kk kk kk
if K is a finite group and L is a subgroup of K then L divides K . Proof . Let 2ß 1 − L . Then if 2B œ 1B, right cancellation yields 2 œ 1. Hence the map is injective. Now assume that C − b. By definition of orbit there is an 2 − L so that C œ 2B, so that the map is also surjective. Therefore, it's a bijection. Since K is finite, L is finite, so that L œ b for each orbit b, since there is a bijection between them. Since the orbits partition K as a set by the previous exercise,
Kœ
-
B− K
bB and the bB are disjoint,
where bB is the orbit of B under the action. Hence,
k k º - º !k k k k k k kk kk K œ
B− K
bB œ
bB œ 8 b œ 8 L
B−K
where 8 is the number of distinct orbits of the action and where the second equality follows from the fact the bB are disjoint. Therefore, by definition L divides K . For the problems below, Geometry: Euclid and Beyond by Robin Hartshorne is a useful very elementary further treatment on geometry. Problem 1.7.20 Show that the group of rigid motions of a tetrahedron is isomorphic to a subgroup of W % .
a b ab
Proof . Denote the group by K. Earlier in the chapter, we showed that a group K acting on a set E induces a permutation a1 − K given by 51 À E Ä Eß 51 + œ 1 † +, and that these permutations in turn induce a homomorphism : À K Ä WE given by : 1 œ 51. Now label the four vertices of the tetrahedron by "ß #ß $ and %. Then if E œ "ß #ß $ß % , we have an induced homomorphism : À K Ä W% given by how each rigid motion 1 − K permutes the vertices of the tetrahedron. By Problem 1.6.13, notice that : K is a subgroup of W %. Moreover, since each rigid motion gives a distinct permutation of the vertices of a tetrahedron (and is completely determined by where it sends the vertices), : is injective, so again by Problem 1.6.13, this means K z : K .
e
f
ab
ab
Problem 1.7.21 Show that the group of rigid motions of a cube is isomorphic to W % .
a b a b e f
Proof . Denote the group by K. Earlier in the chapter, we showed that a group K acting on a set E induces a permutation a1 − K given by 51 À E Ä Eß 51 + œ 1 † +, and that these permutations in turn induce a homomorphism : À K Ä WE given by : 1 œ 51. Now label the four pairs of opposite vertices of the tetrahedron by "ß #ß $ and %. Then if E œ "ß #ß $ß % , we claim K acts on E. First, obviously the identity
