Solutions to Abstract Algebra (Dummit and Foote 3e) Chapter 2 : Subgroups Jason Rosendale
[email protected] February 11, 2012 This work was done as an undergraduate student: if you really don’t understand something in one of these proofs, it is very possible that it doesn’t make sense because it’s wrong. Any questions or corrections can be directed to
[email protected].
Exercise 2.1.1a 0 + 0i is the identity element of G, and a + ai, b + bi ∈ G implies −b + (−b)i ∈ G with (a + ai) + (−b + (−b)i) = (a − b) + (a − b)i ∈ G.
Exercise 2.1.1b 1 is the identity element of G, and a, b ∈ G implies |ab−1 | = |a||b−1 | = |1||1| = 1 so that ab ∈ G.
Exercise 2.1.1c 0 1
is the identity element of G, and a/b, p/q ∈ G implies b|n and q|n so that b = n/x and q = n/y for some x, y ∈ Z. Thus: a p ax py ax − py − = − = b q n n n which, when reduced, becomes =
(ax − py)/ gcd(ax − py, n) n/ gcd(ax − py, n)
The denominator of this fraction clearly divides n, and thus the fraction is ∈ G.
Exercise 2.1.1d 0 1
is the identity element of G, and a/b, p/q ∈ G implies a p aq − bp − = b q bq
And if b, q are relatively prime to n then bq is also relatively prime to n so the fraction is ∈ G.
Exercise 2.1.1e 1 is the identity element of G, and a, b ∈ G implies (ab)2 = a2 b2 which is a product of elements of Q and is thus an element of Q.
1
Exercise 2.1.2a (1 2)(2 3) = (1 2 3), so the set is not closed if n ≥ 3.
Exercise 2.1.2b (sr−1 )(sr) = r2 , so the set is not closed (unless r2 = 1, thus the requirement n ≥ 3).
Exercise 2.1.2c Suppose n is composite. Then we can choose a, b such that ab = n with 1 < a < n. We’re told that there is an element x such that o(x) = n. If {x ∈ G| |x| = n} were a subgroup then the requirement of closure would require that all elements in {1, x, x2 , . . . , xn−1 } be elements of the subgroup. But o(xb ) = a, so xb is not in the set and thus it is not closed under the operation.
Exercise 2.1.2d The operation isn’t closed under addition (1+3 is not odd), and not closed under inverses for multiplication (3−1 = 1/3 6∈ Z).
Exercise 2.1.2e √ √ √ The operation isn’t closed. ( 2 + 3)2 = 5 + 2 6.
Exercise 2.1.3a The identity exists, and each element is its own inverse.
Exercise 2.1.3b The identity exists, r2 is its own inverse, and (sr)−1 = sr3 .
Exercise 2.1.4 Let G = Q − {0} under the operation of multiplication and let H = Z. H is closed under its operation but does not contain inverses.
Exercise 2.1.5 By Langrange’s theorem, |H| = n − 1 must divide n which is not possible when n > 2.
Exercise 2.1.6 Let T represent the torsion subgroup of G. Clearly 1 ∈ T . Suppose now that a, b ∈ T . There must be finite i, j such that ai = bj = 1. From this, we have: 7→ ai = bj = 1 i j
j i
assumed
→ (a ) = (b ) = 1
1i = 1j = 1
→ aij = bij = 1
algebraic substitution
→ aij bij = 1 → (ab)ij = 1 from abelianism To prove that abelianism is necessary, consider the group of bijective functions Z → Z. The torsion group T contains every element of finite order: in particular, it contains the following two bijective functions of order 2: f (x) = x + 1 if x is even , f (x) = x − 1 if x is odd g(x) = x − 1 if x is even , f (x) = x + 1 if x is odd 2
But the composite of these two functions is: (f ◦ g)(x) = x − 2 if x is even , (f ◦ g)(x) = x + 2 if x is odd so that f ◦ g is of infinite order, so that f ◦ g 6∈ T . Thus T is not a group.
Exercise 2.1.7 Note that the operation on Z × (Z/nZ) must be addition in order for it to be a group. Every element of Z/nZ is of finite order, and every nonzero element of Z is of infinite order. So the torsion group must be T = {1} × {Z/nZ} To show that the set of elements of infinite order together with the identity do not form a subgroup of this direct product, we simply take the two elements 3 × 0 and −2 × 1. Their sum, 1 × 1, is a nonidentity element of finite order.
Exercise 2.1.8 Let H ∪ K be a subgroup. Proof by contradiction: Suppose neither H ⊆ K nor K ⊆ H. Then we could find some a ∈ H − K and some b ∈ K − H. But ab 6∈ H, for this would imply a−1 ab = b ∈ H; and ab 6∈ K, for this would imply abb−1 = a ∈ K. Thus ab 6∈ H ∪ K, which contradicts our assumption that H ∪ K was a group.
Exercise 2.1.9 To show that it’s a subgroup of GL, then we can just appeal to basic linear algebra and rely on the facts that det(I)=1 and det(AB)=det(A)det(B).
Exercise 2.1.10a Both H and K contain the identity, so 1 ∈ H ∩ K. If a, b ∈ H ∩ K then a, b−1 ∈ H and a, b−1 ∈ K; thus ab−1 ∈ H, ab−1 ∈ K; therefore ab−1 ∈ H ∩ K.
Exercise 2.1.10b Let AαTbe an arbitrary,T possibly non-countable collection of subgroups of G. All subgroups contain the T identity, so 1 ∈ α Aα . If a, b ∈ α Aα then a, b−1 ∈ Aα for all α; thus ab−1 ∈ Aα for all α; therefore ab−1 ∈ α Aα .
Exercise 2.1.11a This set contains the identity (1,1); and if the set contains (a1 , 1) and (a2 , 1) then it contains (a1 − a2 , 1).
Exercise 2.1.11b This set contains the identity (1,1); and if the set contains (1, b1 ) and (1, b2 ) then it contains (1, b1 − b2 ).
Exercise 2.1.11c This set contains the identity (1,1); and if the set contains (a1 , a1 ) and (a2 , a2 ) then it contains (a1 − a2 , a1 − a2 ).
Exercise 2.1.12a This set contains the identity 1n = 1, and if it contains an , bn then it contains (b−1 )n (from closure under inverses) (ab−1 )n (from abelianism).
Exercise 2.1.12b This set contains the identity 1 since 1n = 1, and if it contains a, b then an = 1, bn = 1 and (bn )−1 = 1, therefore (ab−1 )n = 1 (from abelianism). 3
Exercise 2.1.13 Note: the weird requirement on H just guarantees us that every element of H has a multiplicative inverse even though H is only a group under addition. We prove that if H 6= {0} then H = Q: Suppose we have some nonzero element ab ∈ H. From closure under addition we have b ab = a ∈ H. From the existence of multiplicative inverses we then have a1 ∈ H, and therefore a a1 = 1 ∈ H. Now that we have 1 ∈ H, closure requires n ∈ H for all n ∈ Z and the existence of multiplicative inverses requires n1 ∈ H for all n ∈ Q. And every rational number pq is some combination of these elements, so Q ⊆ H. And since we were told that H ⊆ Q, we have H = Q.
Exercise 2.1.14 The set contains s and sr, but it does not contain ssr = r when n ≥ 3.
Exercise 2.1.15 For ease of notation, let this infinite union be represented by H. Since 1 ∈ H1 , we know that 1 ∈ H. Now suppose a, b ∈ H. There must be some m, n such that a ∈ Hm and b ∈ Hn . So that for N = max(m, n) we have a, b, b−1 ∈ HN and thus ab−1 ∈ HN . This gives us ab−1 ∈ H.
Exercise 2.1.16 The identity matrix fulfills the criteria for the given set (let’s call it U Tn ). Proof by induction that U Tn is a subgroup for all n ∈ N: U T1 , the set of 1 × 1 matrices over F , is trivially a subgroup of GL1 (F ). Assume that U Tk is a subgroup of GLk (F ): we prove that U Tk+1 is also a subgroup. Let A, B be arbitrary elements of U Tk+1 . The inverses are also in U Tk+1 : a11 U Tk (a11 )−1 −U Tk −1 A= , A = 0 ann 0 (ann )−1 And the product AB −1 is in U Tk+1 : a11 U T 1k (b11 )−1 −1 AB = 0 ann 0
−U T 2k (bnn )−1
=
a11 (b11 )−1 0
U T 1k − U T 2k ann (bnn )−1
so that U Tk+1 is a group. By induction, U Tn is a subgroup of GLn (F ) for all n.
Exercise 2.1.17 This proof is almost identical to that of the preceeding exercise. The identity matrix fulfills the criteria for the given set (let’s call it U Ln ). Proof by induction that U Ln is a subgroup for all n ∈ N: U L1 , 1 × 1 matrice over F , is trivially a subgroup of GL1 (F ). Assume that U Lk is a subgroup of GLk (F ): we prove that U Lk+1 is also a subgroup. Let A, B be arbitrary elements of U Lk+1 . The inverses are also in U Lk+1 : 1 U Lk 1 −U Lk −1 A= , A = 0 1 0 1 And the product AB −1 is in U Lk+1 : 1 U L1k 1 −1 AB = 0 1 0
−U L2k 1
=
1 0
U L1k − U L2k 1
so that U Lk+1 is a group. By induction, U Ln is a subgroup of GLn (F ) for all n.
Exercise 2.2.1 CG (A) is a group, so it contains g iff it contains g −1 , so the two sets contain the same elements. More formally: choose g ∈ CG (A). The following steps are all bidirectional (iff):
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7→ g ∈ CG (A) ↔g
−1
assumed
∈ CG (A)
↔ (∀a ∈ A)g
−1
the group CG (A) contains inverses
ag = a
↔ g ∈ {g ∈ G|g
−1
definition of membership in CG (A)
ag = a for all a ∈ A}
Exercise 2.2.2 Members of Z(G) commute with every g ∈ G, and CG (Z) is the set of elements that commute with elements of Z(G): that is, G itself. More formally: 7→ g ∈ CG (Z(G)) ↔ gag
−1
↔ gag
−1
assumed
= a for all a ∈ Z(G)
def. of CG −1
= a for all a such that aga
= g for all g ∈ G
def. of Z
↔ ga = ag for all a such that ag = ga for all g ∈ G
right-multiplication
↔g∈G
last statement is true for all g ∈ G
Exercise 2.2.3 We know that centralizers are groups, so we need only prove that CG (B) ⊆ CG (A). 7→ g ∈ CG (B)
assumed
↔ gbg
−1
= b for all b ∈ B
def. of CG
↔ gbg
−1
= b for all b ∈ A
A⊆B
↔ g ∈ CG (A)
def. of CG
Exercise 2.2.6a Choose h ∈ H. We will prove that h ∈ NG (H) by proving that hHh−1 = H. From the closure of H, we know that hHh−1 ⊆ H. To prove that H ⊆ hHh−1 ⊆ H, we choose h2 ∈ H and note that h(h−1 h2h)h−1 = h2 ∈ hHh−1 . Thus hHh−1 = H and so h ∈ NG (H). We’re also asked to show that (a) does not hold if H is not a subgroup. I find this a bit confusing, since the claim is that H is a subgroup. I’ll assume that we’re really being asked to show that H is necessarily a subset of NG (H) if H is not a subgroup. Consider the group D8 with H = {r, s}. Only the identity element commutes with every element of H, so NG (H) = {1}.
Exercise 2.2.6b This is true practically by definition. By definition, CG (H) is the set of elements that commute with every element of H. So to say H ≤ CG (H) is to say that every element of H commutes with every element of H. More formally: 7→ H is abelian
assumed
↔ (∀h1 , h2 ∈ H)h1 h2 = h2 h1 ↔ (∀h1 , h2 ∈ H)h1 =
h2 h1 h−1 2
def. of abelian right-multiplication
↔ (∀h1 ∈ H)h1 ∈ CG (H)
def. of CG (H)
↔ H ≤ CG (H)
def. of ≤
Each step was bidirectional (iff), so we have proven that H is abelian iff H ≤ CG (H).
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Exercise 2.2.7 Part (a) is a direct consequence of exercise 1.2.5. Part (b) is a direct consequence of exercise 1.2.4.
Exercise 2.2.8 The text of this chapter contains a proof that stabilizers of G are subgroups of G. We could also proceed by defining S = {i} and noting that the kernel of the action of G on S is precisely Gi ; and the kernel of an action is a subgroup.
Exercise 2.2.9 To show NH (A) ⊆ NG (A) ∩ H, choose an arbitrary element h ∈ H: 7→ h ∈ H ∧ h ∈ NH (A)
assumed −1
↔ h ∈ H ∧ (∀a ∈ A)hAh
=A
↔ h ∈ H ∧ h ∈ NG (A)
def. of NH (A) def. of NG (A)
↔ h ∈ H ∩ h ∈ NG (A) Each step was bidirectional, so the two sets are equal.
Exercise 2.2.10 H must contain the identity element and one nonidentity element h: that is, H = {1, h}. We proceed with a bidirectional proof: 7→ g ∈ NG (H)
assumed
↔ (∀h ∈ H)gHg ↔ g1g
−1
−1
=H
= 1 and ghg
↔ (∀h ∈ H)ghg
−1
−1
=h
↔ g ∈ CG (H)
def of NG (H) =h
see below for justification H = {1, h} def. of CG (H)
The middle step is justified in the downward (’only if’) direction by noting that we clearly have g1g −1 = 1, so that if we require gHg −1 = H then it’s necessary that ghg −1 = h. The step is justified in the upward (’if’) direction by noting that {1, h} is a complete list of the elements of H. For the final deduction, we note that if CG (H) = G this means each h ∈ H commutes with every element of G. And, by definition, Z(G) is the set of all elements that commute with every element of G. Therefore H ⊆ Z(G).
Exercise 2.2.11 Choose some z ∈ Z(G). Then zg = gz for all g ∈ G. Therefore zgz −1 = g for all g ∈ G, which means zAz −1 = A for any set A ⊆ G. By definition, this means z ∈ NG (A).
Exercise 2.2.12b The identity action of S4 would be the identity permutation (1). It’s clear from the nature of permutations that (σ1 ◦ σ2 ) · p = σ1 · (σ2 · p).
Exercise 2.2.12c An element of S4 stabilizes 4 iff its disjoint cycles contain the 1-cycle (4). Thus we can construct the ismorphism ϕ : S3 7→ S4 ϕ(σ) = σ ◦ (4)
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Exercise 2.2.12d The permutations that stabilize x1 + x2 are those that either stabilize both 1 and 2 or swap the elements of 1 and 2: (1), (1 2), (1 2)(3 4), (3 4) Showing that this is an abelian group is trivial.
Exercise 2.2.12e The permutations that stabilize x1 x2 + x3 x4 are the above permutations, plus the permutations that swap both 1,3 and 2,4 or 1,4 and 2,3: (1 3)(2 4), (1 4)(2 3), (1 3 2 4), (2 3 1 4) To prove isomorphism, let R = (1324) and S = (1 3)(2 4). We obvious have o(R) = 4, o(S) = 2. And to verify RS = SR−1 : RS = (1 3 2 4)(1 3)(2 4) = (1 2) = (1 3)(2 4)(2 3 1 4) = SR−1 We need now only prove that ϕ(S i Ri S j Rj ) = si ri sj rj is a homomorphism from this set of permutations to D8 . And this follows immediately from the associativity of D8 : ϕ(S i Ri S j Rj ) = si ri sj rj = (si ri )(sj rj ) = ϕ(S i Ri )ϕ(S j Rj )
Exercise 2.2.12f The same logic applies: the permutations that stabilize (x1 + x2 )(x3 + x4 ) are those permutations that swap elements in exactly the same way as in part (e).
Exercise 2.2.13 The proof is identical to that given for n = in the preceeding exercise.
Exercise 2.2.14 It’s not specified if we’re looking for the center of H(F ) in H(F ) or the center of H(F ) is GL3 (F ). We’ll assume the more general case for now and look for the center of H(F ) in GL3 (F ). H(F ) is just the set of 3 × 3 upper-triangular matrices with diagonal elements of 1. The center, Z(H(F )), consists of the elements A ∈ GL3 (F ) such that AH = HA for all H ∈ H(F ). Calculating AH and HA: 1 x y a ax + b ay + bz + c a b c AH = d e f 0 1 z = d dx + e dy + ez + f g h i 0 0 1 g gx + h gy + hz + i
1 HA = 0 0
x 1 0
y a z d 1 g
b e h
c a + dx + gy f = d + gz i g
We need to see when these two matrices are equal for all H, so we’re we must have d = g = h = 0, so we simplify to a ax + b ay + bz + c a , HA = 0 ez + f AH = 0 e 0 0 i 0 So that we must have a = e = i. Further simplifying: a ax + b ay + bz + c , az + f AH = 0 a 0 0 a
a HA = 0 0
7
b + ex + hy e + hz h
c + f x + iy f + iz i
solving for A. We see immediately that ex + b e 0
iy + f x + c iz + f i
ax + b a 0
ay + f x + c az + f a
So that our final requirement is that bz = f x. The only way for this to hold for all values of x, z is for b = f = 0. Therefore, the center of H(F ) is the set of matrices of the form a 0 c A= 0 a 0 0 0 a Now we see that the exercise must have been asking us to find the center of H(F ) in H(F ), for there is not necessarily any isomorphism between F and the set of all such Z(H(F )) (consider F = Z2 : |F | = 2 but |Z(H(F )))| = 4). Restricting our center to H(F ) gives us 1 0 c A= 0 1 0 0 0 1 which has an obvious isomorphism with F :
ϕ : F → H(F ),
1 ϕ(x) = 0 0
0 1 0
x 0 1
Exercise 2.3.2 If o(x) = n, then x0 , x1 , . . . , xn−1 are distinct by exercise 1.1.32. Thus | hxi | = |G| and hxi ⊆ G, and they’re both finite sets so they must contain the same elements. To show that this does not hold for infinite groups, let G = Z and let x = 2.
Exercise 2.3.3 Any n such that gcd(n, 48) = 1 (i.e., numbers with no factors of 2 or 3).
Exercise 2.3.4 Any n such that gcd(n, 202) = 1 (i.e., numbers with no factors of 2 or 101).
Exercise 2.3.5 It’s tempting to resort to Euler’s Phi function and say the answer is ϕ(49000). To find this number explicitly, we first find how many integers have factors in common with 49000. Using the theorem from set theory that |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|, we have (factors of 2 + factors of 5 + factors of 7) − (factors of 2,5 + factors of 5,7 + factors of 2,7) + (factors of 2,5,70) = (24500 + 9800 + 7000) − (4900 + 1400 + 3500) + (700) = 32200 So that ϕ(49000) = 4900 − 32200 = 16800.
Exercise 2.3.8 Any a such that gcd(a, 48) = 1 (i.e., numbers with no factors of 2 or 3).
Exercise 2.3.9 We’ll use the additive notation instead of the multiplicative notation and find the integers a such that ψa : 1 7→ ax (i.e., ψa (k) = kax) extends to a well-defined homomorphism. We’ll see that ϕa is well-defined only iff 3|a. We prove that well-definedness requires this. Suppose psia is a well-defined homomorphism. Choose i = j ∈ Z:
8
7→ i = j
given: equality in equality in Z/48Z
→ i = j + 48
equality in Z
→ ψa (i) = ψa (j + 48)
ψa is well-defined: equality in Z36
→ ψa (i) = ψa (j + 48) + 36n
ψa is well-defined: equality in Z
→ ψa (i) − ψa (j) = ψa (12) + ψa (36) + 36n)
ψ is a homomorphism: equality in Z
the operations on the groups Z and Z/48Z are addition, so xa is repeated addition. ψ(kx) = kψ(x) : → ψa (i) − ψa (j) = 12ψa (1) + 36(ψa (1) + n)
Thus
ψ is a homomorphism: equality in Z
Moving to Z3 6, the LHS of this equation is zero because ψ is well defined and 36k = 0 in Z36 : → 12ψa (1) = 0
ψa is well-defined: equality in Z36
→ (∃n ∈ Z)12ψa (1) = 36n
equality in Z
→ (∃n ∈ Z)psia (1) = 3n → 3|ψa (1)
equality in Z def. of divisibility
→ 3|a
def. of ψa
This really just shows that ψ is a well-defined homomorphism only if 3|a. To prove the if part, we assume a = 3k and define ψa : 1 7→ 3kx. We need only prove well-definedness, since homomorphism follows trivially. 7→ i = j
assumed: equality in equality in Z/48Z
→ i = j + 48n
equality in Z
→ ψa (i) − ψa (j + 48n) = 3kxi − 3kx(j + 48)
def of ψa : equality in Z36
→ ψa (i) − ψa (j + 48n) = 3kxi − 3kx(j + 48) + 36m
equality in Z36
→ ψa (i) − ψa (j + 48n) = 3kx(i − j) + 3(48)kx + 36m
algebraic rearrangment: equality in Z
→ ψa (i) − ψa (j + 48n) = 3kx(i − j) + 36(4kx + m)
algebraic rearrangment: equality in Z
→ ψa (i) − ψa (j + 48n) = 3kx(i − j)
equality in Z36
→ ψa (i) − ψa (j + 48n) = ψa (i − j)
def. of ψa : equality in Z36
→ ψa (i) − ψa (j) = ψa (0)
in Z/48Z we have j = j + 48 and i = j: equality in Z3 6
→ ψa (i) − ψa (j) = 0
def of ψa : equality in Z3 6
→ ψa (i) = ψa (j)
algebraic rearrangement: equality in Z3 6
Exercise 2.3.10 The order is lcm(30, 54)/30 = 54/ gcd(30, 54) = 9
Exercise 2.3.11 A subgroup that is not cyclic is {1, r2 , s, sr2 }. Each element has order 1 or 2.
Exercise 2.3.12a We can prove this by enumeration. There are only four elements of Z2 × Z2 , each of which has order 1 or 2.
Exercise 2.3.12b Proof by contradiction. Suppose the element (a, b) generated Z2 × Z2 . Then (a + 1, b) cannot be an element of h(a, b)i for this would imply that n(a, b) = (a + 1, b) for some n which would mean that n = 1 and thus a = a + 1, which is not true for any element of Z2 .
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Exercise 2.3.12c Proof by contradiction. Suppose the element (a, b) generated Z × Z. Then (a + 1, b) cannot be an element of h(a, b)i for this would imply that n(a, b) = (a + 1, b) for some n which would mean that n = 1 and thus a = a + 1, which is not true for any element of Z.
Exercise 2.3.13 For parts (a) and (b), the group on the left-hand side contains an element (0, 1) of order 2 while the right-hand side contains no element of order 2.
Exercise 2.3.15 Proof by contradiction. Suppose the element (a, b) generated Q × Q. Then (a + 1, b) cannot be an element of h(a, b)i for this would imply that n(a, b) = (a + 1, b) for some n which would mean that n = 1 and thus a = a + 1, which is not true for any element of Q.
Exercise 2.3.16 We’re told that o(x) = n, o(y) = m. Let A = lcm(m, n). From the relation mn = gcd(m, n)lcm(m, n) we have (xy)A = xA y A m A/m
= (x )
from commutativity n A/n
(y )
A/m, A/n are both integers
=1 Thus o(xy)|A. This does not necessarily hold if x, y do not commute (consider D8 with o(r) = 4, o(s) = 2, but o(sr) 6= 4). The easiest example of commuting elements such that o(xy) is not equal to lcm(x, y) is to take D8 with x = y = r. We have o(rr) = 2 while lcm(o(r), o(r)) = lcm(4, 4) = 4.
Exercise 2.3.17 hx|xn = 1i
Exercise 2.3.18 Choose x ∈ Zn such that hxi = Zn . Define ϕ : Zn → H as ϕ : xk → hk . To show that this is a homomorphism is trivial. To show uniqueness, assume we had another homomorphism f with f (x) = h. Then we have, from homomorphism: f (xk ) = f (x)k = hk = ϕ(xk ) so that f = ϕ.
Exercise 2.3.19 Define ϕ : Z → H as ϕ(n) = hn . This is a homomorphism: ϕ(a + b) = ha+b = ha hb = ϕ(a)ϕ(b) And this function is unique: if f is any homomorphism with f (1) = h, we have (using multiplicative notation so that 1n = n): f (n) = f (1n ) = f (1)n = hn = ϕ(n)
Exercise 2.3.20 n
If xp = 1, then we know that o(x)|pn . But p is prime, so the only divisors of pn are p0 , p1 , p2 , . . . , pn . Thus o(x) = pm for some m ≤ n.
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Exercise 2.3.21 From the binomial theorem we have pn−1
(1 + p)
=
n−1 pX
k=0
n−1 pX n k n−k n k p 1 = p k k
k=0
We want to find the value of this term mod pn : n−1 pX
k=0
pn−1 k p mod pn k
Any terms of pk with k ≥ n are going to become zero here, so this is equivalent to =
n n−1 X p k=0
k
pk mod pn
n−1 The coefficient p k will be a multiple of pn−1 whenever pn−1 > k, which can (with a bit of algebra) be shown to hold for all n, k whenever p > 2. Thus every term in this sum except the k = 0 term has a factor of pn and we can simplify: 0 n−1 X p = pk = 1 mod pn k
k=0
It’s not enough to show that this doesn’t always hold for pn−2 ; if we want to make the deduction they’re asking us to make, it must *never* hold.
Exercise 2.3.24a If g ∈ NG (hxi), then g hxi g −1 = hxi, which means that gxg −1 ∈ hxi, and therefore gxg −1 = xa for some a ∈ Z.
Exercise 2.3.24b Note that (gxg −1 )n = g(xg −1 g)n−1 xg −1 = gxn g −1 So suppose gxg −1 = xa . Then we have gxn g −1 = (gxg −1 )n = (xa )n . This is sufficient to prove that g hxi g −1 ⊆ hxi. Since G is a finite group, o(x) = n is finite. For i, j = 0, 1, . . . , n − 1 we have ⇐⇒ gxi g −1 = gxj g −1 ⇐⇒ gxi g −1 gx−j g −1 = 1 ⇐⇒ xi−j = g −1 g = 1 ⇐⇒ i = j So that gxi g −1 is distinct for each i = 0, 1, . . . , n − 1. Thus |g hxi g −1 | = | hxi |. Since these are finite subsets with g hxi g −1 ⊆ hxi, this proves that g hxi g −1 = hxi.
Exercise 2.3.25a Suppose that G = hxi with |G| = o(x) = n. Let k be an integer relatively prime to n. We prove that o(xk ) = n: 7→ o(xk ) = i
G is finite, so this must be true for some i > 0
k i
→ (x ) = 1 → xki = 1 → o(x)|ki → n|ki
o(x) = n
→ n|i
k is relatively prime to n
k The
k least i > 0 such that n|i is obviously i = n, so o(x ) = n. Thus the size of the range of this map is | x | = n = |G|, which for finite groups is sufficient to prove xk = G.
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Exercise 2.3.25b Let G be a finite group with |G| = n (not necessarily cyclic). Choose k to be relatively prime to n. Define the mapping f : G → G as f (g) = g k . We’re asked to show that this mapping is surjective. Choose an arbitrary element g ∈ G. Each element has finite order, so let o(g) = m. Since m|n (by Langragne’s theorem), we know that k is relatively prime to m. Thus, applying the previous exercise, f (hgi) = hgi and thus g is in the range of f . But g was arbitrary, so every g ∈ G is in the range of f . And thus f is surjective.
Exercise 2.3.26a We proved this in exercise 25. We proved the ’if’ portion explicitly, and we were assured that the ’only if’ portion is a direct consequence of Cauchy’s Theorem which will be presented in section 3.2.
Exercise 2.3.26b σa (x) = σb (x) ⇐⇒ xa = xb ⇐⇒ xa−b = 1 ⇐⇒ (a − b)|n ⇐⇒ a = b(mod n)
Exercise 2.3.26c Let Zn = hxi. Suppose ϕ : Zn → Zn is an automorphism. Every element of Zn is expressible as xi for some i, so ϕ(x) = xi for some i. And this completely defines ϕ, since for each xj ∈ hxi we have ϕ(xj ) = ϕ(x)j = (xi )xj .
Exercise 2.3.26d σa ◦ σb (x) = σa (σb (x)) = σa (xb ) = xab = σab (x)
Exercise 2.4.1 hHi is defined to be the intersection ∩Ai of all subgroups Ai such that H ⊆ Ai ≤ G. Since H ∈ {Ai } we have ∩Ai ⊆ H; since H ⊆ Ai for each Ai , we have H ⊆ ∩Ai . Thus H = ∩Ai = hHi.
Exercise 2.4.2 If x ∈ hAi then x is the finite product of elements of A; since A ⊆ B, this means that x is the finite product of elements of B; thus x ∈ hBi. To see that we can have A ⊂ B with hAi = hBi, consider the additive group Z with A = {1} and B = {1, 2}.
Exercise 2.4.3 Each element of H commutes with every element of H (by abelianism) and every element of Z(G) (by definition). In turn, every element of Z(G) commutes with every element of H (since H ⊂ G) and every element of Z(G) (since Z(G) ⊂ G). This means that all the elements from H, Z(G) commute with one another. Thus hH, Z(G)i is abelian. However, the elements of CG (H) do not necessarily commute with other elements of CG (H). Consider G = GLn , H = In . H is just the identity element, so it’s trivially abelian. Every element commutes with the identity element, so we have CG (H) = G. Thus hH, CG (H)i = G. But we know that G is not abelian.
Exercise 2.4.4 First, suppose H = {1}. Then hH − {1}i = h∅i. This represents the intersection of every subgroup A such that ∅ ⊆ A ≤ G. But the empty set is a subset of every set, so h∅i is the intersection of every subgroup of G; the only commonality is 1. Thus hH − {1}i = {1} = hHi. Now suppose H contains some nonidentity element h. Then 1 = hh−1 ∈ hH − {1}i, so that hH − {1}i = hHi = H.
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Exercise 2.4.5 The only elements of order 2 in S3 are the two 2-cycles (c.f. exercise 1.4.13). Without loss of generality, we can write these as (a b) and (b c). It’s trivial (but tedious) to construct six distinct elements from finite products of these two elements.
Exercise 2.4.6 This is equivalent to h(1 2), (3 4)i. From exercise 1.4.13, we know that any product of commuting 2-cycles has order 2.
Exercise 2.4.7 We let R = AB = (1 3 2 4) and S = A = (1 2). Define the isomorphism ϕ : D8 → S4 as r 7→ R, s 7→ S. We need only show that the relations from the generator ho(r) = 4, o(s) = 2, sr = r−1 si hold in S4 . And this is done easily, since o(R) = 4, o(S) = 2, and SR = R−1 S = B.
Exercise 2.4.8 Let A = (1 2 3 4), B = (1 2 4 3). Then hA, Bi contains A−1 = (1 4 3 2) and A−1 BA−1 = (1 2). On page 64, we’re assured that the two elements A, (1 2) generate all of S4 (although we’re told that the proof will come in a later chapter). If this proof isn’t convincing, we can always hunker down and generate the 24 distinct elements of S4 from finite products of A, B.
Exercise 2.4.9 In a previous exercise, we’ve shown that |GL2 (Fp )| = p4 − p3 − p2 + p. In this case, that gives us |GL2 (F3 )| = 48. Half of these have determinant 1, and half have determinant 2. Thus |SL2 (F3 )| = 24. We haven’t encountered any groups that are isomorphic to this group, so I don’t see any way to prove that these two elements generate SL2 (F3 ) other than actually generating all 24 distinct elements.
Exercise 2.4.10 In exercise 1.5.3, we developed a generator for Q8 : Q8 = h1, i, j, k|i2 = j 2 = k 2 = −1, ij = ki Let the two given elements of SL2 (F3 ) be represented by I, J and define the following: 0 -1 1 1 -1 1 1 0 I= , J= , K = IJ , 1= 1 0 1 -1 1 1 0 1 Remember that in F3 we have 2 = −1. It’s easy to verify that o(i) = o(j) = o(k) = 4, and that i2 = j 2 = k 2 = −1. From section 1.6, this mapping between the generators of Q8 and the generators of whatever we’re calling this subset of SL2 (F3 ) is sufficient to prove isomorphism.
Exercise 2.4.11 The subgroup described above has only one element of order 2, while S4 has more than one (e.g., (1 2) and (3 4)). Thus the two groups cannot be isomorphic by exercise 1.7.2.
Exercise 2.4.12 This isn’t true unless we restrict ourselves to upper-triangular matrices with 1s along the diagonal (cf exercise 2.1.17). In this case, all such matrices are of the form 1 a b 0 1 c 0 0 1 13
where a, b, c ∈ {0, 1}. There are clearly 23 = 8 1 R= 0 0
such matrices. Now, 1 0 1 1 1 , S = 0 0 1 0
then we have o(R) = 4, o(S) = 2. Furthermore, we have 1 1 1 1 R−1 = R3 = 0 1 1 , SR = 0 0 0 1 0
define the matrices R, S to be 0 0 1 1 0 1
1 1 0
0 0 = R−1 S 1
at which point we’ve reproduced the generator for D8 .
Exercise 2.4.13 Let P = {1/p|p is prime }. Clearly hP i ⊆ Q+ . Now, choose r/qQ∈ Q+ . From the fundamental theorem of algebra, we know that q is the finite product of prime powers q = pni i . Therefore we have Y 1 1 r ∈ hP i = r Q ni = r pi q pni i so that Q+ ⊆ hP i.
Exercise 2.4.14a Every finite G is generated by itself: G = hGi.
Exercise 2.4.14b The additive group Z is generated by h1i.
Exercise 2.4.14c Let H be a finitely generated subgroup and let A be the finite set of generators for H such that hAi = H. Define k, 1/k as described in the exercise. It’s clear that h1/ki generates each element in A and therefore generates all of H. Thus H ⊆ h1/ki.
Exercise 2.4.14d If Q were finitely generated, then it would itself be a finitely generated subgroup of Q, and would therefore be cyclic by part (c). But it’s not cyclic. By contrapositive, then, Q can not be finitely generated.
Exercise 2.4.15 Consider the additive group of the set of powers of 2: i 2 |i ∈ Z
This set cannot be generated by 2k , because the set generated by this element will never yield 2k−1 (remember: operation is addition).
Exercise 2.4.16a If |G| is finite, then there are only (|G| − |H|) elements in G − H, and therefore there are precisely (2|G|−|H| − 1) proper subsets of G containing H. If none of these finitely many subsets is a maximal group, then that means that H was its own maximal group in G.
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Exercise 2.4.16b By Lagrange’s theorem, the order of any subgroup of D8 must divide 8. So the only subgroup of D8 with order > 4 must be D8 itself.
Exercise 2.4.16c Suppose H ≤ G = hxi and |G| = o(x) = n. Each element of G, and thus H, is of the form xi for some i ∈ {0, . . . , n − 1}. To satisfy group closure, we must have H = hxa i where a = gcd({i|xi ∈ H}). We now consider three cases: 1. If (a, n) = 1 then H = G and H is not maximal (c.f. proposition 6 in section 2.3). 2. If a|n and a = m1 m2 is composite we see that H = hxa i ≤ hxm1 i ≤ G so that H is not maximal. 3. If a|n and a = p is prime, then hxp i is maximal. For suppose we have a group H 0 containing H and some additional element xi so that H 0 = xp , xi . If p|i then xi is already in H and thus H = H 0 . If p does not divide i, then the primality of p means (p, i) = 1 and thus H 0 = G. In either case, no proper subgroup of G contains H as a proper subgroup, so H is maximal. We don’t need the case where (a, n) > 1 but a 6 |n, since in this case we would have x(a,n) ∈ hxa i so
to consider that hxa i = x( a, n) and clearly (a, n)|n. Therefore these three cases exhaust all possibilities and prove that H is maximal iff H = hxp i for some prime p.
Exercise 2.4.17a −1 By the subgroup criterion, we need only show that for S each a, b ∈ H we also have ab ∈ H. Define H to be the union of chain elements H1 , H2 , . . . so that H = Hi . Then we have:
a, b ∈ H → (∃i, j ∈ N)a ∈ Hi , b ∈ Hj → (∃j ∈ N)ab−1 ∈ Hj → ab−1 ∈ H
Exercise 2.4.17b Let {g1 , g2 , . . . , gn } represent the finite generators of G. From our definition of S, each element in chain C must be a proper subgroup of G. So if H = G then for each gi we could find ki ∈ N such that gi ∈ Hki . Then, since this set is finite, we could find the maximum of {ki }: call it k. Thus each gi ∈ Hk , and so hg1 , g2 , . . . , gn i ≤ Hk , thus G = Hk . And this contradicts our claim that each Hi is a proper subgroup of G.
Exercise 2.4.17c We’re pretty much done. We have a bunch of chains in S, all of which have an upper bound (by exercise 2.4.16(a)); thus by Zorn’s Lemma S contains a maximal element.
Exercise 2.4.18a n
i
Fix p prime, and let Z = {z ∈ C|z p = 1 for some n ∈ Z+ }. Define Hi = {z ∈ Z|z p = 1}. We have a bidirectional proof showing that Hk ≤ Hm ⇐⇒ m ≥ k.
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7→ Hk ≤ Hm
assumed
↔ z ∈ Hk → z ∈ Hm k
m
↔ zp = 1 → zp = 1 k
m
↔ (z p )a = z p , a > 0
(note: a ∈ N k
To justify the claim that a > 0, we note that for z p = 1 to always imply z p m k z p be a positive integer power of z p .
m
= 1 we must have
↔ apk = pm , a > 0 To justify this last step in the → direction, note that we must have a = b if we want z a = z b for all z ∈ Z. ↔ a > 0, a = pm−k ↔m≥k
Exercise 2.4.18b From the given assumption, we can immediately conclude that D E n e2πim/p = Hk
Exercise 2.4.18c n
Let Y < Z. For each y ∈ Y , we have some smallest n such that y p = 1 so that y = exp(2πim/pn ) for some m. Our choice of the smallest such n allows us to know that (m, p) = 1 (for otherwise (m, p) = pk and we have n−k y = exp(2πi(mp−k )/pn−k ) so that y p = 1). From part (b), this means that hyi contains n distinct elements including exp(2πi/pn ). The closure of Y then assures us that exp(2πi/pn ) ∈ Y . Now now consider the set of all such exponentials generated by elements y ∈ Y , and further consider the set of powers of p in this set: 2πi ∈ Y n ∈ N |exp pn S∞ For every n in this set, we have Hn ⊆ Y . Thus, if this set has no upper bound, then Y = n=1 Hk = Z and Y is S not a proper subgroup. If this set does have an upper bound, then it has a least upper bound N and thus Y = n=1 N − 1Hk = HN −1 . So we have proven that Y is a proper subgroup iff Y = Hk for some k ∈ Z+ .
Exercise 2.4.18d Let gi1 , . . . , gin be an arbitrary finite set of elements of Z. We prove that this cannot be a generator of Z. Using the logic of the previous exercise, for each gi we can find a subgroup Hj such that gi ∈ Hj . By exercise 2.4.2 this implies hgi1 i ≤ hHj1 i. Thus by part (a) we have hgi1 , . . . , gin i ≤ hHj1 , . . . , Hjn i = Hj , j = max(j1 , . . . , jn ) And clearly Hj 6= Z, so Z was not generated by this finite set.
Exercise 2.4.19a For each p/q ∈ Q and each integer k we have p ∈ Q, k qk
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p qk
=
p q
Exercise 2.4.19b Let A be a nontrivial finite abelian group. Every element in a finite group has a finite order. So define k to be the finite product Y k= o(ai ) ai ∈A
Since A is nontrivial, we can find some nonidentity element b ∈ A. But this element has no k-th root: by our definition of k we see that o(a)|k for all a ∈ A and thus ak = 1 6= b for all a ∈ A.
Exercise 2.4.20 First, assume that A × B is divisible. Then all of the elements of the form (1A , b) and (a, 1B ) are divisible, which can only happen if A and B are both divisible. Next, assume that A and B are divisible: let (a, b) be an arbitrary element of A × B and let k be an arbitrary element of Z+ . By divisibility we have a1/k ∈ A, b1/k ∈ B and thus (a1/k , b1/k ) = (a, b)1/k ∈ A × B
Exercise 2.5.1 through 2.5.20 These exercises would require way too much time to typeset and I’m kind of eager to move on to quotient groups. You’re on your own.
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