Solutions to Abstract Algebra (Dummit and Foote 3e) Chapter 1 : Group Theory 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 1.1.1 (a) is not associate: 3 − (2 − 1) 6= (3 − 2) − 1. (b),(c),and (d) can be shown to be associative by grinding out the algebra and showing that a ∗ (b ∗ c) = (a ∗ b) ∗ c. (e) is not associative, as can be shown by letting a = b = c = 2.
Exercise 1.1.2 (a) and (e) can be shown to not be commutative by letting a = 1, b = 2. (b),(c) and (d) can be shown to be commutative by grinding out the algebra.
Exercise 1.1.3 a + b+ c = a+ b+ c
def. of modular addition
= a+ b + c
def. of modular addition
Exercise 1.1.4 ab · c = a· b· c
def. of modular multiplication
= a · bc
def. of modular multiplication
Exercise 1.1.5 To be a group, each element a would need a multiplicative inverse b such that ab = 1; but 0 has no such inverse.
Exercise 1.1.6 (b) is not closed: 1/14 + 1/14 = 1/7. (c) is not closed: 2/3 + 2/3 = 4/3. (d) is not closed: 2 + (−3/2) = 1/2. (f ) is not closed: 1/2 + 1/3 = 5/6. (a) and (e) are groups under addition.
Exercise 1.1.7 That x?y is well-defined follows directly from the fact that +,-, and the greatest integer function are well defined on R. To prove that G is a group: G has an identity element (0 ∈ G and 0 ? x = x ? 0 = x − [x] = 0), each
1
element a ∈ G has an inverse (1 − a ∈ G and a ? (1 − a) = 1 − [1] = 0 = (1 − a) ? a), and associativity can be shown with a little tedious algebra. That G is abelian follows from the commutativity of addition: x ? y = (x + y) − [x + y] = (y + x) − [y + x] = y ? x
Exercise 1.1.8a G has an identity element (1 ∈ G and z1 = z = 1z) and an inverse (z n = 1 implies (z −1 )n = 1 so z −1 ∈ G) and complex multiplication is associative. To show closure: 7→ x, y ∈ G
assumed +
→ (∃m, n ∈ Z )x +
→ (∃mn ∈ Z )x
m
mn
n
=1∧y =1 n
=1 =1∧y
def. of membership in G mn
m
=1
=1
→ (∃mn ∈ Z+ )xmn y mn = 1 → (∃mn ∈ Z+ )(xy)mn = 1 → xy ∈ G
def. of membership in G
Exercise 1.1.8b The group is not closed under addition. 1 and −1 are both elements of G, but 1 + (−1) = 0 is not.
Exercise 1.1.9a The identity is (0, 0), the inverse of (a, b) is (−a, −b), associativity and closure are trivial.
Exercise 1.1.9b The √ identity is (1, 0), associativity and closure are trivial. The inverse of (a, b) can be found by solving (a + b 2)x = 1 and is given by: a b (a, b)−1 = , a2 − 2b2 a2 − 2b2
Exercise 1.1.10 The definition of a symmetric matrix is that for all i, j we have aij = aji . In the group table, this is the case iff for all i, j we have ai aj = aj ai which is the definition of abelian.
Exercise 1.1.11 General formula is: |a| = lcm(a, 12)/a. |0| = 1 |1| = 12 |2| = 6 |3| = 4 |4| = 3 |5| = 12 |6| = 2 |7| = 12 |8| = 3 |9| = 4 |10| = 6 |11| = 12
2
Exercise 1.1.12 |1| = 1 | − 1| = 2 |7| = 2 | − 7| = 2 |13| = 2
Exercise 1.1.13 General formula is: |a| = lcm(a, 36)/a. |1| = 38 |2| = 18 |6| = 6 |9| = 4 |10| = 18 |12| = 3 | − 1| = 36 | − 10| = 18 | − 18| = 2
Exercise 1.1.14 This would be tedious to type out. For each a we need to use the methods of exercises 0.3.15 to find y such that 36x + ay = 1 and this y would be the order of a.
Exercise 1.1.15 Proof by induction. For the n = 1 case, it’s clear that a1 = a−1 1 . Supposing the equality holds for the n = k case, we have: −1 −1 (a1 a2 . . . ak ak+1 )(a−1 k+1 ak . . . a1 ) −1 −1 = (a1 a2 . . . ak )ak+1 a−1 k+1 (ak . . . a1 )
= =
−1 (a1 a2 . . . ak )1(a−1 k . . . a1 ) −1 (a1 a2 . . . ak )(a−1 k . . . a1 )
=e −1 (a−1 k+1 ak
associativity definition of inverses definition of identity equality holds for n = k
. . . a−1 1 )
and thus is shown to be the right inverse of (a1 a2 . . . ak ak+1 ). The proof for left inverse is similar. Thus equality holds for the n = k + 1 case. By induction, equality holds for all n ∈ Z+ .
Exercise 1.1.16 This follows directly from the definition of |x|. The order of x is the smallest integer n such that xn = 1; if x2 = 1, then this smallest integer must be less than or equal to 2. Thus it can be only 1 or 2.
Exercise 1.1.17 If |x| = n, then xn = 1, and thus x · xn−1 = 1, which makes xn−1 the inverse of x.
3
Exercise 1.1.18 7→ xy = yx
assumed
→y
−1
−1
xy = y
→y
−1
xy = 1x = x
yx
left multiplication definition of inverse
→ x−1 y −1 xy = x−1 x →x
−1 −1
→ (x
y
left multiplication
xy = 1
−1 −1 −1
y
)
definition of inverse
= xy
definition of inverse
→ xy = yx
consequence of exercise 1.1.15
Exercise 1.1.19a Part (a) is a direct consequence of associativity.
Exercise 1.1.19a Part (b) is a special case of exercise 1.1.15.
Exercise 1.1.19c Parts (a) and (b) trivially hold when a = 0 or b = 0. Parts (a) and (b) can be shown to hold for a, b ≤ 0 by taking the inverses of each side of each equation. The only thing left to show is that part (a) holds when a < 0 < b or b < 0 < a: Assume that a < 0 < b or b < 0 < a. If |a| = |b| (absolute value, not order) then clearly xa+b = x0 = xa x−a = xa xb ow assume, without loss of generality, that |a| < |b| Then b has the same sign as −a and b has the same sign as a + b. Therefore −a has the same sign as a + b and we can deduce: xa xb = xa x−a+b+a
from b = b + a − a
= xa x−a xb+a
from part (a), since −a and b + a have same sign
= (xb+a
left cancellation
Exercise 1.1.20 From 1.1.19(b), we have xn = 1 ↔ (xn )−1 = 1−1 = 1 ↔ x−n = 1 ↔ (x−1 )n = 1 Thus the least n such that xn = 1 must also be the least n such that (x−1 )n = 1.
Exercise 1.1.21 7→ x2k+1 = 1
assumed, since n = 2k + 1
→ x2k+1 x = 1x = x
right multiplication
→ x2k+2 = x
1.1.19(a)
2 k+1
→ (x )
=x
1.1.19(a)
4
Exercise 1.1.22 7→ xn = 1 ↔ (xgg
assumed
−1 n
gg −1 = 1
) =1
↔ xg(g −1 xg)n−1 g −1 = 1
tricky associativity
↔ g −1 xg(g −1 xg)n−1 g −1 g = g −1 1g = 1
left and right multiplication
↔ g −1 xg(g −1 xg)n−1 = 1
right cancellation
↔ (g
−1
n
xg) = 1
1.1.19(a)
This shows that the least n such that x = 1 must also be the least n such that (g −1 xg)n = 1. To prove that |ab| = |ba|, let x = ab and g = b−1 in the preceeding proof and conclude that the least n such that (ab)n = 1 must also be the least n such that (babb−1 )n = (ba)n = 1. n
Exercise 1.1.23 (xs )t = xst = xn = 1, so |xs | ≤ t. But if there were some k such that 0 ≤ k < t such that (xs )k = 1, then we would have 0 ≤ sk < st = n with xsk = 1 contradicting the assumption that |x| = n.
Exercise 1.1.24 Proof by induction. The n = 0 and n = 1 cases are trivial. Assume the equality holds for n = k: (ab)k+1 = a(ba)k b (= a(ab)k b k k
(= aa b b
tricky associativity a, b commute equality holds for n = k
k+1 k+1
(= a b Thus equality holds for all n ∈ N. To prove that this holds for negative n, just take the inverse of each side of the equation to yield (ab)−n = b−n a−n , then apply commutativity to conclude (ba)−n = b−n a−n .
Exercise 1.1.25 Given that x2 = 1, we see that (ab)(ab) = b(aa)b = 1. Right-multiplying by b−1 a−1 yields ab = ba.
Exercise 1.1.26 We’re told that H is nonempty so it contains some element h; it’s closed under inverses and the binary operation, it contains the identity element hh−1 = 1. It inherits associativity from G. That’s sufficient for H to be a group.
Exercise 1.1.27 Let H = xn : n ∈ Z. The identity x0 = 1 exists in n; if xn ∈ H then x−n ∈ H so H is closed under inverses; it’s trivially closed under the binary operation; and (xi xj )xk = xi+j+k = xi (xj xk ) so it is associative.
Exercise 1.1.28 All parts of this exercise can be proven through simple but tedious algebra.
Exercise 1.1.29 7→ (a, b) · (c, d) = (c, d) · (a, b) ↔ (ac, bd) = (ca, db)
def. of ×
↔ ac = ca and bd = db
def. of equality in A × B
5
Exercise 1.1.30 To prove commutativity: (a, 1)(1, b) = (a1, 1b) = (a, b) = (1a, b1) = (1, b)(a, 1) Now let p = |a|, q = |b|, and x = [p, q]. Then: (a, b)x = (ax , bx ) = ((ap )x/p , (bq )x/q ) = (1x/p , 1x/q ) = (1, 1) = 1 Thus the order of (a, b) divides [p, q] = x. Now let n = |(a, b)|. From exercise 1.1.24: (1, 1) = (a, b)n = [(a, 1)(1, b)]n = (a, 1)n (1, b)n Right-multiplying gives us: (1, b)−n = (a, 1)n which means (1, b−n ) = (an , 1) thus an = b−n = 1. Thus p and q both divide n, which means that [p, q] divides n. We’ve now shown that n = |(a, b)| divides x = [p, q] and that x divides n: therefore x = n.
Exercise 1.1.31 To prove that t(G) has an even number of elements, establish an equivalence relation a b : (ab = 1ora = b) on t(G). This is clearly reflexive,symmetric, and easily shown to be transitive. There are two elements per equivalence class on t(G): one element and its inverse. Thus t(G) has an even number of elements. Thus the set G − t(G) (that is, g ∈ G|g = g −1 ) must also have an even number of elements. It contains the identity, so it must contain at least one nonidentity element. Each element in G − t(G) has g 2 = 1, and therefore must have order 1 or 2 by exercise 1.1.16. Only the identity has order 1, so this nonidentity element must have order 2.
Exercise 1.1.32 Suppose xi = xj for 0 ≤ i < j ≤ n − 1. Then 1 = xj x−i = xj−i , which would contradict the claim that |x| = n.
Exercise 1.1.33a Part (a) is the contrapositive of exercise 1.1.31.
Exercise 1.1.33b It’s trivial to show that i = k implies xi xi = x2k 1 and therefore xi = x−1 . To prove in the other direction: 7→ xi = x−i assumed ↔ xi xi = x−i xi ↔ x2i = 1 This shows that xi = x−i iff n (the order of x) divides 2i. But 2i < n by the previous exercise, so n|2i implies 2i = 0 or 2i = n. And we’re asked to assume that i > 0, so we must have 2i = n = 2k and so i = k.
Exercise 1.1.34 Were it the case that xm = xn for some m 6= n, then we would have x|m−n| = 1 which would contradict the assumption that x was of infinite order.
Exercise 1.1.35 Let x ∈ G have order n. Let k be an arbitrary power of x. By the division algorithm, we can find unique values of a, b with 0 ≤ b < n such that k = an + b. Thus we have xk = xan+b = xa nxb = xb 6
Exercise 1.1.36 We know that at least one element must be its own inverse from exercise 1.1.31. WLOG, assume that this element is a. From 1.1.31 we know that b is its own inverse iff c is its own inverse: if both are their own inverses, then the group is abelian by 1.1.25 and we are done. If neither b nor c are their own inverses, then they must have order 3 (b3 = c3 = 1). We cannot have b2 = b (for we would have o(b) = 1) or b2 = 1 (for we would have o(b) = 2) or b2 = a (for we would have b = b, b2 = a, b3 = ab, b4 = a2 = 1 so that o(b) = 4) so we must have b2 = c. And this gives us a contradiction, since it would imply that o(b2 ) = o(c) = o(b).
Exercise 1.2.2 We know that D2n has order 2n with distinct elements r0 , r1 , . . . rn−1 , r0 s, r1 s, . . . rn−1 s. So any element that is not equivalent to rk for some k is equivalent to srk for some k. We can now use a proof by induction: the case for x = r0 s and x = r1 s is trivial. Now assume that the equality holds for x = rk s: r(rk+1 )s = rr(rk )s = r(rk )sr−1 = rk+1 sr−1 and thus equality holds for x = rk+1 s.
Exercise 1.2.3 As above, every element of D2n which is not a power of r has the form rk s = sr−k . Thus: (rk s)2 = (rk s)(rk s) = sr−k rk s = ss = 1 This proves only that the order of (rk s) divides 2; we must show that it is 2. Proof by contradiction: If the order were 1, then we would have rk s = 1 which would imply s = r−k which would imply that all 2n elements could be written uniquely as a power of r; but rn = 1 so there could be at most n such elements. Thus the order of (rk s) cannot be 1. To show that s, sr are generators is suffices to show that ssr = r, so all powers of r can be generated.
Exercise 1.2.4 z 2 = r2k = rn = 1, so o(z) = 2 (k > 1, so z is not the identity element). r2 k trivially commutes with elements of D2n that are powers of r; since o(z) = 2 it commutes with all other elements of order 2 (cf 1.1.33); that is, all elements of D2n that are not a power of r (previous exercise). To show that this is the only element other than the identity that commutes with all elemenets of D2n , we use a very unsatisfying proof by exhaustion of cases. Let a be an element such that ab = ba for all b ∈ D2n . Either a = ri or a = sri for some i ∈ 0 . . . 2k − 1. case i) Suppose a = ri and let b be an arbitrary element of D2n . The element a obviously commutes for all b of the form b = rj . If b = srj , then ab = ba when: 7→ ab = ba i
j
assumed j i
↔ r sr = sr r −i j
definition of a, b
↔ sr r = sr r
from relation rs = sr−1
↔ r−i+j = rj+i
left cancellation
↔r
−i+j −j−i
r
j i
=1
right multiplication by r−(j+i)
↔ r−2i = 1 ↔ n|2i
o(r) = n
↔ k|i
n = 2k
This shows that, for this choice of a = ri , ab = ba for all b iff i is a multiple of k. But we’ve assumed that i ∈ 0, . . . , 2k − 1 so either i = 0 or i = k. Thus a = ri commutes with all b iff a = r0 (the identity) or a = rk . 7
case ii) Suppose a = sri . Then this a does not commute with all b: let b = r. Then we have 7→ ab = ba
assumed
i
i
definition of a, b
i+1
−1+i
from relation rs = sr−1
↔ sr r = rsr ↔ sr
= sr
↔ ri+1 = ri−1
left cancellation
↔ ri+1 r1−i = 1
right multiplication by r−(−j+i)
↔ r2 = 1 But n is defined by the order of r, so r2 = 1 implies n = 2, contradicting the requirement that n ≥ 4. Thus we have shown that the only nonidentity element that commutes with all elements of D2n (n = 2k, n ≥ 4) is rk .
Exercise 1.2.5 As shown in the previous exercise, the only element that commutes with element of D2n for n > 2 is ri where n|2i. If n is odd then the requirements that i ∈ 0, . . . , n and i = kn together imply i = 0 and thus the only commuting element is r0 = 1.
Exercise 1.2.6 Since o(x) = o(y) = 2, we have x = x−1 , y = y −1 . Thus we have tx = xyx tx = xy
right-multiply t = xy by x −1 −1
x
y = y −1 , x = x−1
tx = x(xy)−1 tx = xt−1
Exercise 1.2.7 Let a = s and b = sr so that ab = r. The relations follow from each other: a2 = s2 = 1, (ab)n = rn = 1, and b2 = (sr)2 = ssr−1 r = 1.
Exercise 1.2.8 n is, by definition, the smallest r such that rn = 1 so o(r) = n.
Exercise 1.2.9 through 1.2.13 A tetrahedron has four three-sided faces; a cube has six four-sided faces; an octahedron has eight three-sided faces; a dodecahedron has twelve five-sided faces; an icosahedron has twenty three-sided faces. Each side (pair of vertices) can be flipped, so the total number of positions for a side is : The number of rigid motions for a given solid is given by: 2 ∗ (number of sides) =
(number of faces) ∗ (number of sides per face) number of faces connected to each side
Each solid has two faces connected to each side, so this reduces to (number of faces) * (number of sides per face).
Exercise 1.2.14 For addition, 1. For multiplication, p : p is prime (by the fundamental theorem of arithmetic).
Exercise 1.2.15 For addition, p : (p, n) = 1. For multiplication, p < n : p is prime. 8
Exercise 1.2.16 Letting n = 2, we have r2 = s2 = 1. So let x = r, y = s. The relations follow: x2 = r2 = 1,y 2 = s2 = 1, and xy = y −1 x : xy = (xy)−1 =y
from o(xy) = 2
−1 −1
x
= yx−1
from o(y) = 2
Exercise 1.2.17a The presentation in 1.2 is hx, y|xn = y 2 = 1, xy = yx2 i. The text demonstrated that xy = yx2 implies x3 = 1. Thus, if n = 3k then x3 = 1 but x0 , x1 , x2 are distinct elements. Letting x = r, y = s gives us x3 = r3 = 1, y 2 = s2 = 1, and rs = sr−1 : rs = xy
from o(xy) = 2
2
= yx
= yx−1
xx2 = 1 → x2 = x−1
= sr−1
Exercise 1.2.17b We know that x3 = 1 so x3k = 1. By definition, xn =1. If (3, n) = 1 then either n = 3k + 1 or n = 3k + 2 = 3(k + 1) − 1 for some k. In either case, we have 1 = xn = x3k±1 = x3k x±1 = x±1 But x−1 = 1 iff x = 1, so in either case we have x = 1 and X2n is generated uniquely by s, which has order 2.
Exercise 1.2.18 The presentation in 1.3 is hu, v|u4 = v 3 = 1, uv = v 2 u2 i. a) If v 3 = vv 2 = 1, then v 2 = v −1 . Similarly, u2 = u−2 and u3 = u−1 . b)
u3 = u(v 3 )u2 2 2
= (uv)(v u ) 2 2
from v 3 = 1 associative property
= (v u )(uv)
from uv = v 2 u2
= v 2 u3 v
associative property
=v
−1 3
u v
from part (a)
Left-multiplying both sides by v gives us vu3 = u3 v; right-multiplying by v −1 gives us u3 v −1 = v −1 u3 . c) From part (b), we have u3 v −1 = v −1 u3 . Using part (a) this becomes u−1 v −1 = v −1 u−1 . Taking inverses of each side yields vu = uv. d) We’re told uv = v 2 u2 ; by commutativity of u, v this becomes uv = (uv)2 ; by left- or right-cancellation this becomes uv = 1. e) 1 = (uv)4 = u4 v 3 v = v, and uv = 1 implies u = v −1 = 1.
Exercise 1.3.8 Define the function f : N → Sω to be f (n) = (1n). The function is clearly injective (albeit not surjective) so |Sω | ≥ |N|.
Exercise 1.3.9 See exercise 1.3.11. 9
Exercise 1.3.10 Trivial proof by induction on i. Since σ m maps ak → ak , it must be the identity function; since m is the least integer such that σ m = I, we have o(σ) = m.
Exercise 1.3.11 Let k be the order of σ i . It must be the case that m|ik or, by the division algorithm, we could find ik such that ik = xm + b with 0 < b < m which would give us (σ i )k = σ ik = σ xm+b = (σ m )x σ b = σ b which would contradict the previous exercise’s conclusion that o(σ) = m. Having proven that ik must be a multiple of m, we see that the least such multiple of ik is ik = lcm(m, ik). If we want the o(σ i ) = m, we must have im = lcm(m, im) = lcm(m, i) which occurs only when gcd(m, i) = 1: ab = lcm(a, b) · gcd(a, b) → lcm(a, b) =
ab gcd(a, b)
Exercise 1.3.12a yes: let σ = (1 3 5 7 9 2 4 6 8 10) and k = 5.
Exercise 1.3.12b no. Let k be the smallest k such that σ k = (1 2)(3 4 5). Then σ 2k = (3 5 4). Thus σ 2k (1) = 1 (implying that m|2k from exercise 10) but σ 2k (3) = 5 (implying m - 2k by exercise 10). This establishes a contradiction so there can be no k such that σ k = τ .
Exercise 1.3.13 Assuming that σ can be written has a product of commuting 2-cycles, we have σ 2 = [(a1 b1 ) . . . (ak bk )]2 = (a1 b1 )2 . . . [(ak bk )2 = (1)2 . . . (1)2 = (1) so that o(σ) = 2. If we assume that o(σ) = 2 then σ maps a → σ(a) and maps σ(a) → σ(σ(a)) = a (with the possiblity that σ(a) = a). Consider the set of 2-cycles: {(a σ(a)) : a ∈ 1, . . . , m and a < σ(a)} We can define σ as the product of every element of this set: σ = (a1 σ(a1 ))(a2 σ(a2 )) . . . (ak σ(ak )) Each a ∈ 1, . . . , m appears in at most one of these disjoint 2-cycles, and appears iff a 6= σ(a).
Exercise 1.3.14 Follow the preceeding proof. Assuming that σ can be written has a product of commuting p-cycles, we have σ p = [(a1 b1 ) . . . (ak bk )]p = (a1 b1 )p . . . [(ak bk )p = (1)p . . . (1)p = (1) And we can write σ as the product of elements of the disjoint collection of p-cycles: {(a σ 1 (a) σ 2 (a) . . . σ p−1 (a)) : a ∈ 1, . . . , m and a = min(a, σ(a), . . . , σ m−1 (a))} If p is not prime then we can find a, b > 1 such that p = ab = lcm(a, b) and the element (1 2 . . . a)(a + 1 a + 2 . . . a + b) has order p despite not being a product of disjoint p-cycles. A more explicit example: In S6 we have (1 2 3)(4 5) which has order 6. 10
Exercise 1.3.15 Choose σ ∈ Sn and let k be the least common multiple of the lengths of the cycles in its cycle decomposition. From exercise 1.1.24 we know that σ k = (1) means that (a1 . . . ai )k = (1) for each cycle in the decomposition of σ, so k must be a common multiple of the order (length) of each cycle in the decomposition of σ; thus the order of σ must be the least such k, which is the least common multiple of the lengths of the cycles of σ.
Exercise 1.3.16 n From n elements, there are m ways of selecting m elements and m! ways of writing an m−cycle with them. Each distinct m-cycle can be written in m different ways. Thus the number of distinct m-cycles in Sn is given by n!m! n(n − 1)(n − 2) . . . (n − m + 1) n m! = = m!(n − m)!m m m m
Exercise 1.3.17 There are n4 ways of selecting the elements of two disjoint 2-cycles, and 3 unique ways to construct the two 2-cycles with them. Thus the number of distinct products of two disjoitn 2-cycles is: n! 1 n(n − 1)(n − 2)(n − 3) n 1 = = (n − 4)!4! 3 8 4 3
Exercise 1.3.18 For each Sn we can choose i, k > 0 such that i + k ≤ n and construct following disjoint product Sn : (1 2 . . . i)(n n − 1 . . . n − (k − 1)) By exercise 15, this element has order of lcm(i, k). Thus for S5 we can construct elements of orders 1,2,3,4,5 (trivially) and 6. The order 6 element is given by (1 2 3)(4 5)
Exercise 1.3.19 With the explanation from the previous exercise we see that S7 has elements of order 1, 2, 3, 4, 5, 6, 7 (trivially), lcm(2, 5) = 10, and lcm(3, 4) = 12.
Exercise 1.3.20 Let a = (1 2), b = (2 3) and define the generator to be ha, b|a2 = b2 = 1, abab = bai. All other elements of S3 can be written with these two elements: (1) = a2 = b2 (1 3) = aba = bab (1 2 3) = ab (1 3 2) = ba The fact that a2 = b2 = 1 means that every element of S3 can be reduced to an alternating string of a, b. The choice of the relation abab = ba (and its equivalents, ab = baba and aba = bab) means that any such alternating string of length n ≥ 4 can be reduced to a string of length n − 2. Thus S3 can be represented as string of 3 or fewer alternating elements a, b. There are only 6 such strings: 1, a, b, ab, ba, and aba = bab.
Exercise 1.4.1 Proof by enumeration. Consider all 16 possible 2 × 2 matrices with entries in {0, 1} and show that exactly 6 of them have nonzero determinants.
11
Exercise 1.4.2
0 1
2
1 0 1 0
1 0
= 0 1
1
0 1
2
1 1 1 1
1 1
= 3
=
1 1
0 1
1 1
2
0 1 1 0
=I 3 =I
Exercise 1.4.3
1 1
0 1
1 0
1 1
=
1 1
1 0
6=
=
1 0
1 1
1 1
0 1
Exercise 1.4.4 Suppose n is not prime, and let a(1 < a < n) be a divisor of n. Then a has no multiplicative inverse. Proof by contradiction: assume that ak = 1. Then we would have ak + mn = 1 (by definition of equivalence mod n) which means that gcd(a, n) = 1 which contradicts our assumption that a(1 < a < n) is a divisor of n.
Exercise 1.4.5 2
If |F | = q is finite, then there are at most q n possible n × n matrices; GLn (F ) is a subset of these matrices 2 and at least one such matrix has a zero determinant, so |GLn (F )| ≤ q n and thus GLn (F ) is finite. If |F | is infinite, then f I ∈ GLn (F ) for each f ∈ F (where I is the identity matrix) and thus |GLn (F )| ≥ |F | which means |GLn (F )| is infinite.
Exercise 1.4.6 see previous exercise
Exercise 1.4.7 The determinant of a 2 × 2 matrix is given by the formula a b det = ad − bc c d So we see that the deterimant is zero if ad = bc. Using basic combinatorics, it’s easy to show that: • We can choose a, d such that ad = 0 in p + (p − 1) distinct ways • We can choose a, d such that ad = 1 in p − 1 distinct ways • We can choose a, d such that ad = 2 in p − 1 distinct ways • ... • We can choose a, d such that ad = p − 1 in p − 1 distinct ways The same holds true for the number of ways we can choose b, c. Which means • We can choose a, b, c, d such that ad = bc = 0 in (p + (p − 1))2 distinct ways • We can choose a, b, c, d such that ad = bc = 1 in (p − 1)2 distinct ways • We can choose a, b, c, d such that ad = bc = 2 in (p − 1)2 distinct ways • ... • We can choose a, b, c, d such that ad = bc = p − 1 in (p − 1)2 distinct ways
12
Thus there are (2p − 1)2 ways to have ad and bc equal to 0, and (p − 1)2 ways to have them equal each of the (p − 1) other values. Thus the total number of ways we can construct a 2 × 2 matrix with ad = bc is (2p − 1)2 + (p − 1)(p − 1)2 = p3 + p2 − p And since there are p4 possible 2 × 2 matrices over Fp , the total number of such matrices with nonzero determinants is p4 − p3 − p2 + p
Exercise 1.4.8 Let A be the matrix with a1,2 = 1 as the only nonzero entry, and let B be the matrix with a2,1 as the only nonzero entry. Then AB has a1,1 = 1 as the only nonzero entry while BA has a2,2 as the only nonzero entry.
Exercise 1.4.9 We want to show that a1 b1 a2 c1 d1 c2
b2 d2
a3 c3
b3 d3
b2 c2
a1 c1
a1 a2 0
=
b1 d1
a2 c2
b2 d2
a3 c3
b3 d3
This can be done tediously through algebra.
Exercise 1.4.10a
a1 0
b1 c1
a2 0
=
a 1 b2 + b1 c 2 c1 c2
Exercise 1.4.10b We want to find values of a2 , b2 , and c2 such that the product in part (a) is the identity matrix. It’s immediately −1 clear that we need to have a2 = a−1 1 and c2 = c1 . With these substitutions, we have a1 b2 + b1 c2 = a1 b2 + b1 c−1 1 −1 which equals 1 exactly when b2 = a−1 1 (1 − b1 c1 ). So the inverse is −1 −1 a1 a−1 1 (1 − b1 c1 ) 0 c−1 1 −1 This is an element of G since a−1 1 6= 0, c1 6= 0.
Exercise 1.4.10c G is a group: we’ve shown closure under the operation in part (a), closure of inverses in part (b), associativity in exercise 9, and it’s clear that the identity matrix is an element of G. We’re told that a 6= 0 and c 6= 0, so all elements of G have a nonzero determinant ac − 0b = ac.
Exercise 1.4.10d Parts (a) through (c) are still valid after adding the further restriction that a1 = c1 . The proof changes very little.
13
Exercise 1.4.11a
1 a b 1 XY = 0 1 c 0 0 0 1 0 To prove non-abelianism, we see calculate Y X: 1 d e 1 Y X = 0 1 f 0 0 0 1 0
d 1 0
e 1 f = 0 1 0
d+a 1 0
e + af + b f +c 1
a 1 0
b 1 c = 0 1 0
a+d 1 0
b + dc + e c+f 1
So we have XY 6= Y X whenever af 6= cd. An explicit example can be given by letting a = b = f = 0, c = d = e = 1.
Exercise 1.4.11b We want to find values of d, e, and f such that both of the products in part (a) are the identity matrix. This immediately yields a system of equations: d+a
=
0
e + af + b =
0
b + dc + e
=
0
c+f
=
0
Whose solution is d = −a, f = −c, e = ac − b. So the inverse matrix is 1 −a ac − b −c X −1 = 0 1 0 0 1
Exercise 1.4.11c Associativity can be proven with tedious algebra. The previous parts of this exercise show that H(F ) is a group. The fact that each of the 3 entries can take |F | possible values implies that o(H(F )) = |F |3 .
Exercise 1.4.11d Too tedious to typeset.
Exercise 1.4.11e Let X be an arbitrary element of H(R). We prove by induction that for all n ∈ N, the matrix X n has the form ac 1 na nb + n(n−1) 2 Xn = 0 1 nc 0 0 1 The case for k = 1 is trivial. For k = 2 we have 2 1 a b 1 X2 = 0 1 c = 0 0 0 1 0 Now assume that we have established the 1 a b 1 X k+1 = XX k = 0 1 c = 0 0 0 1 0
2a 1 0
2b + ac 2c 1
form of X k . For X k+1 : ka kb + k(k−1) ac 1 2 = 0 1 kc 0 1 0
(k + 1)a 1 0
(k + 1)b + (k + 1)c 1
k(k+1) 2
ac
So that the proof by induction is complete. From this, we see that X n = I only if na = 0, nc = 0, and nb + n(n + 1)ac = 0. This occurs only when a = b = c = 0: that is, when X is the identity matrix. Thus every nonidentity element has infinite order. 14
Exercise 1.5.1 o(1) = 1, o(−1) = 2, o(±i) = o(±j) = o(±k) = 4.
Exercise 1.5.3 All the given relations of Q8 can be derived from h1, i, j, k|i2 = j 2 = k 2 = −1, ij = ki: ij = k → iij = ik → −j = ik, i = −kj ij = k → −kij = 1 → −ij = −k → j = ki, −i = kj
Exercise 1.6.1 Trivial proof by induction on n. It’s trivial for n = 1, and true by definition of homomorphism for n = 2. Assuming it holds for n = k, we have ϕ(xk+1 ) = ϕ(xk x) = ϕ(xk )ϕ(x) = ϕ(x)k ϕ(x) = ϕ(x)k+1 From this, we have 1 = ϕ(x0 ) = ϕ(x−n xn ) = ϕ(x−n )ϕ(xn ) = phi(x−n )ϕ(x)n so that, by the definition of inverses, −1
ϕ(x−n ) = (ϕ(x)n )
= ϕ(x)−n
Exercise 1.6.2 lemma From the previous exercise, we know that ϕ(x0 ) = ϕ(x)0 so that ϕ(1G ) = 1H . Now, choose an arbitrary x ∈ G and suppose o(x) = n is finite. 7→ xn = 1G n
→ ϕ(x ) = ϕ(1G ) n
→ ϕ(x) = 1H
assumed ϕ is well-defined from previous exercise
→ o(ϕ(x))|n → o(ϕ(x))|o(x)
definition of n
Note that this last step also implies that o(ϕ(x)) is finite. Now assume that o(ϕ(x)) = m is finite: 7 ϕ(x)m = 1H → → ϕ(xm ) = ϕ(1G )
assumed ϕ is well-defined
→ xm = 1G
ϕ is an isomorphism, thus 1-to-1
→ o(x)|m → o(x)|o(ϕ(x))
definition of m
This last step implies that o(x) is finite. Thus we have shown that o(x) is finite iff oϕ(x) is finite, and if either is finite then o(x) = o(ϕ(x)). The result is not true if ϕ is only assumed to be a homormorphism (the step requiring isomorphism is clearly labeled). As a counter example, consider the following homormorphism : f : Z → Z defined as f (n) = 1. f (3)f (2) = 1 · 1 = 1 = f (6), but clearly o(3) = ∞ while o(f (3)) = 1.
Exercise 1.6.3 7→ G is abelian
assumed
↔ (∀a, b ∈ G)ab = ba
def. of abelianism
↔ (∀a, b ∈ G)ϕ(ab) = ϕ(ba) ↔ (∀a, b ∈ G)ϕ(a)ϕ(b) = ϕ(b)ϕ(a)
isomorphisms are well-defined and bijective def. of homomorphisms
↔ ϕ(G) is abelian
def. of abelianism
↔ H is abelian isomorphisms are surjective Each step in this proof is bidirectional, so we’ve proven that when ϕ is isomorphic, then G is abelian iff H is 15
abelian. If ϕ is only a homomorphism, then we have the unidirectional proof: 7→ G is abelian
assumed
→ (∀a, b ∈ G)ab = ba
def. of abelianism
→ (∀a, b ∈ G)ϕ(ab) = ϕ(ba)
isomorphisms are well-defined
→ (∀a, b ∈ G)ϕ(a)ϕ(b) = ϕ(b)ϕ(a)
def. of homomorphisms
→ ϕ(G) is abelian
def. of abelianism
This shows that when ϕ is a homormorphism, then G is abelian implies ϕ(G) is abelian. Note that nothing can be assumed from the abelianism of H or ϕ(G). Consider ϕ : G → 1H defined as ϕ(g) = 1H . ϕ(G) = H is trivially abelian, but G can be any group whatsoever.
Exercise 1.6.4 C − {0} has an element of order 4 (i), but no such element exists in R. This contradicts exercise 1.6.2.
Exercise 1.6.5 Proof 1: There can be no bijective function between the two sets, as proven by Cantor’s Theorem. Proof 2: suppose ϕ : Q → R were √ an isomorphism. Let x ∈ R be the element such that ϕ(2) = x; let a ∈ Q be the element such that ϕ(a) = k. From this, we have √ ϕ(a2 ) = ϕ(a)2 = ( k)2 = k = ϕ(2) √ Which, since ϕ is a bijection, means that a2 = 2 and thus a = 2. But this can’t be true of any a ∈ Q.
Exercise 1.6.6 Let Q has an element of order 3 ( 13 ), but there is no such element in Z. This contradicts exercise 1.6.2.
Exercise 1.6.7 D8 has only one element of order 4 (o(r) = 4) while Q8 has three such elements (i, j, k). This contradicts exercise 1.6.2.
Exercise 1.6.8 Assuming m, n > 0, Sm contains m! elements while Sn contains n!, so there can be no bijection between them unless n = m.
Exercise 1.6.9 D24 has an element of order 12 (r) while S4 can have no such element (exercise 1.3.15)
Exercise 1.6.10a 7→ a = b → θ−1 (a) = θ−1 (b) → σ◦θ
−1
(a) = σ ◦ θ
→ θ◦σ◦θ
−1
θ is bijective −1
(b)
(a) = θ ◦ σ ◦ θ
permutations are well-defined −1
(b)
permutations are well-defined
Exercise 1.6.10b The same logic used in part (a) can show that θ ◦ σ −1 ◦ θ−1 is well-defined, and this clearly acts as an inverse to ϕ.
16
Exercise 1.6.10c ϕ(σ ◦ τ ) = θ ◦ (σ ◦ τ ) ◦ θ−1 = θ ◦ (σ ◦ θ−1 ◦ θ ◦ τ ) ◦ θ−1 = ϕ(σ) ◦ ϕ(τ )
Exercise 1.6.11 The function f : A × B → B × A defined as f (a, b) = (b, a) is easily shown to be an isomorphism.
Exercise 1.6.12 The function f : (A × B) × C → A × (B × C) defined as f ( ((a, b), c) ) = (a, (b, c)) is easily shown to be an isomorphism.
Exercise 1.6.13 Verifying the group properties is tedious but easy. We just need to show that 1H ∈ ϕ(G) and that ϕ(G) is closed under its operation and inverses. If ϕ is injective then the homomorphism is bijective, since ϕ is clearly surjective onto ϕ(G), and bijective homomorphisms are isomorphisms.
Exercise 1.6.14 Verifying the group properties of K is tedious but easy. It’s clear that ϕ can’t be injective if more than one element maps to 1 ∈ H, so ϕ is injective only if K = {1G }. Proof by contrapositive that ϕ is injective if K = {1G }: Assume that K = {1G }. 7→ ϕ(a) = ϕ(b) → ϕ(a)ϕ(b)−1 = 1H −1
→ ϕ(a)ϕ(b
ϕ(G) is a group by previous exercise
) = 1H
→ ϕ(ab−1 ) = 1H → ab−1 = 1G
K = {1G }
→a=b
Exercise 1.6.15 (x, y) is in the kernel of π if π(x, y) = x = 1, so the kernel is K = {(x, y) ∈ R2 |x = 1} = {1} × R
Exercise 1.6.16 Following the logic above, we see that the kernel of π1 is K = {(a, b) ∈ A × B | a = 1A } = {1A } × B and the kernel of π2 is K = {(a, b) ∈ A × B | b = 1B } = A × {1B }
Exercise 1.6.17 Let ϕ : G → G be defined by ϕ(g) = g −1 . This function is clearly onto, so ϕ(G) = G. So we are asked to prove that ϕ is a homomorphism iff G is abelian. First assume that ϕ is a homomorphism: 7→ ϕ(a−1 b−1 ) = ϕ(a−1 )ϕ(b−1 ) −1
→ ϕ((ba)
→ ba = ab
−1
) = ϕ(a
−1
)ϕ(b
)
assumed properties of inverses definition of ϕ
Thus ϕ(G) = G is abelian. Now assume that G is abelian: 17
7→ b−1 a−1 = a−1 b−1 −1
→ (ab)
−1 −1
=a
b
assumed property of inverses
→ ϕ(ab) = ϕ(a)ϕ(b) definition of ϕ Thus ϕ is a homomorphism.
Exercise 1.6.18 Let ϕ : G → G be defined by ϕ(g) = g 2 . 7→ ba = ab ↔ a(ba)b = a(ab)b ↔ a(ba)b = a(ab)b
left- and right-multiplication
↔ (ab)(ab) = (aa)(bb)
associativity
↔ ϕ(ab) = ϕ(a)ϕ(b)
definition of ϕ
Exercise 1.6.19 As definied here, G is the set of finite roots of unity in C. From complex analysis, we know that for each k ∈ N there are k distinct elements of order k. Let k be fixed and define fk as fk (z) = z k . This is clearly a homomorphism, and is surjective since z ∈ G → z 1/k ∈ G → fk (z 1/k ) = z But by exercise 14, fk cannot be injective: the k roots of unity of order k mean that the kernel of fk is of size k.
Exercise 1.6.20 The identity element is the identity mapping; isomorphisms are invertible and therefore have inverses in Aut(G). Associativity and closure is inherited from the properties of function composition.
Exercise 1.6.21 Define fk : Q → Q to be fk (q) = kq. This function is clearly injective and a homomorphism. To prove surjectivity: q q q ∈ Q → ∈ Q → fk =q k k
Exercise 1.6.22 Define fk : A → A to be fk (a) = ak . Since A is abelian, we have fk (ab) = (ab)k = ak bk = fk (a)fk (b) If k = −1, then the function is injective (a = b → a−1 = b−1 → f (a) = f (b)) and surjective (a ∈ A → f (a−1 ) = a).
Exercise 1.6.23 Let σ be an automorphism such that σ 2 is the identity map and σ(g) = g iff g = 1. Choose arbitrary elements a, b ∈ G. σ is bijective, so there are x, y ∈ G such that σ(x) = a, σ(y) = b. From this we have 7→ σ(xy)2 = xy = σ(x)2 σ(y)2
σ 2 is the identity map
→ σ(x)σ(y)σ(x)σ(y) = σ(x)σ(x)σ(y)σ(y)
σ is a homomorphism
→ abab = aabb
definition of a, b
→ ba = ab left- and right- cancellation of previous step and since a, b were arbitrary this suffices to prove that G is abelian. 18
Exercise 1.6.24 We need to show that mapping preserves the properties of each generator and relation. We’re told that x2 = r2 = 1 and y 2 = s2 = 1. From the exercise 1.2.6, the fact that x2 = y 2 = 1 is sufficient to conclude that xy = yx−1 .
Exercise 1.6.25a Let ~v = |v| cos(φ) + |v| sin(φ) be an arbitrary vector. The given matrix transforms ~v as follows: cos(θ) − sin(θ) |v| sin(φ) |v|[cos(φ) cos(θ) − sin(φ) sin(θ)] = sin(θ) cos(θ) |v| cos(φ) |v|[sin(φ) cos(θ) + cos(φ) sin(θ)] which, via the angle addition formulas from the trigonmetric identities, is equivalent to |v| cos(φ + θ) |v| sin(φ + θ) which is, of course, the original vector with its endpoint rotated by an additional φ radians counterclockwise about the origin. Our vector ~v was arbitrary, so every vector endpoint (and thus every point in R2 ) is also rotated in the same way.
Exercise 1.6.25b We need to show that mapping preserves the properties of each generator and relation. ϕ represents a rotation of 2π/n radians, so clearly o(ϕ(r)) = n. And ϕ(s)2 = I, so o(ϕ(s)) = 2. We can show that the relationship rs = sr−1 has an associated relationship ϕ(r)ϕ(s) = ϕ(s)ϕ(r)−1 by some tedious algebraic verification. The text (bottom of p38) assures us that this is sufficient to guarantee an isomorphism between G and D2n .
Exercise 1.6.26 Further define ϕ as follows: ϕ(1) = I, ϕ(k) =
0√ − −1
√ − −1 0
We need to show that mapping preserves the properties of each generator and relation. There are a lot of relations that need to be algebraically verified (e.g., ϕ(i)2 = −ϕ(1), ϕ(i)ϕ(j) = k) but they are trivial (albeit tedious). The identity is the only element of Q8 that maps to I ∈ GL2 (C), so by exercise 1.6.14, this is sufficient to prove that ϕ is injective.
Exercise 1.7.1 Let F be a field and define a group action of G = F × on A = F by g · a = ga. The element 1 ∈ F × satisfies property (i) of group actions (1 · a = a for all a ∈ F ). To prove property (ii), we note that (F, ·) is not a group (0 has no inverse), but it is still a semigroup (it’s associative and closed under its operation). Property (ii) is then justified as follows: (g1 g2 ) · a = g1 g2 a
definition of the group action; F is closed under multiplication
= g1 (g2 a)
multiplication in F is associative
= g1 (g2 · a)
definition of group action
= g1 · (g2 · a)
definition of group action
19
Exercise 1.7.2 The element 0 ∈ Z satisfies property (i) of group actions (0 · a = a for all a ∈ Z). To justify property (ii), let g1 , g2 , a be arbitrary elements of Z: (g1 + g2 ) · a = (g1 + g2 ) + a
definition of the group action; Z is closed under addition
= g1 + (g2 + a)
addition in Z is associative
= g1 + (g2 · a)
definition of group action
= g1 · (g2 · a)
definition of group action
Exercise 1.7.3 The element 0 ∈ R satisfies property (i) of group actions. To justify property (ii), let r1 , r2 be arbitrary elements of R and let (x, y) be an arbitrary point in R × R: (r1 r2 ) · (x, y) = (r1 + r2 ) · (x, y)
operation on R is addition
= (x + (r1 + r2 )y, y)
definition of group action
= (x + (r1 + r2 )y, y)
operation on R is addition
= ((x + r1 y) + r2 y, y
associativity, distributive property of field R × R
= r2 · (x + r1 y, y)
definition of group action
= r2 · (r1 · (x, y))
definition of group action
= g1 + (g2 + a)
addition in Z is associative
= g1 + (g2 · a)
definition of group action
= g1 · (g2 · a)
definition of group action
Exercise 1.7.4a Let K represent the kernel of the action (g ∈ G : g · a = a for all a ∈ A). By the subgroup criterion (proven in chapter 2), we need show that a left identity exists and that a, b ∈ K → ab−1 ∈ K. It’s clear that 1 ∈ K, so an identity exists. Now assume that g1 , g2 ∈ K: 7→ 1, g1 , g2 ∈ K
assumed
→ (∀a ∈ A)g1 · a = a ∧ g2 · a = a ∧ 1 · a = a
definition of K
→ (∀a ∈ A)g1 · (g2 · a) = a
algebraic substitution
→ (∀a ∈ A)(g1 g2 ) · a = a
property (ii) of group actions
→ g1 g2 ∈ K
definition of K
Exercise 1.7.4b Fix some a ∈ A and let S represent the stabilizer of a in G. By the subgroup criterion (proven in chapter 2), we need show that a left identity exists and that a, b ∈ S → ab−1 ∈ S. It’s clear that 1 ∈ S, so an identity exists. Now assume that g1 , g2 ∈ S:
20
7→ 1, g1 , g2 ∈ S
assumed
→ g1 · a = a ∧ g2 · a = a ∧ 1 · a = a
definition of S
→ g1 · (g2 · a) = a
algebraic substitution
→ (g1 g2 ) · a = a
property (ii) of group actions
→ g1 g2 ∈ S
definition of S
Exercise 1.7.5 Each step in the following proof is bidrectional (iff): 7→ g ∈ K ↔ (∀a ∈ A)ga = a
assumed definition of K
↔ (∀a ∈ A)σg (a) = a
definition of σg
↔ σg is the identity permutation on G
definition of the identity function
↔ σg is the identity element of SA
definition of the group SA
↔ g is in the kernel of ϕ : G → SA
definition of kernel, ϕ
Exercise 1.7.6 Proof by contradiction. Assume that G is not faithful: then there are distinct nonidentity elements g1 , g2 such that g1 · a = g2 · a for all a ∈ A. From this, we obtain (g1−1 g2 ) · a = g1−1 · (g2 · a) = g1−1 · (g1 · a) = (g1−1 g1 ) · a = 1 · a = a and thus g1−1 g2 ∈ K. And this element cannot be the identity since g1 6= g2 . Thus the kernel contains a nonidentity element. By contrapositive, if K = {1} then G is faithful.
Exercise 1.7.7 The kernel of the given action is {1}; by the previous exercise, this suffices to prove that the action is faithful.
Exercise 1.7.8a Let G = SA and let B = P(A). The identity permutation σ1 satisfies property (i) of group actions (σ1 (b) = b for all b ∈ B). To show that property (ii) is satisfied, let σg , σh be arbitrary elements of G and let b be an arbitrary element of B: (σg ◦ σh ) · b = (σg ◦ σh )(b)
definition of ·
= σg (σh (b))
definition of ◦
= σg · (σh · b)
definition of ·
Exercise 1.7.8b The element (1 2) acts on each subset by replacing 1 (if it exists) with 2 and vice-versa. For example, (1 2){1, 4} = {2, 4}. The element (1 2 3) replaces each 1 with 2, each 2 with 3, and each 3 with 1. For example, (1 2 3){2, 3, 4} = {3, 1, 4}.
Exercise 1.7.9 The proof in 1.7.8(a) and the description in 1.7.8(b) still apply when subsets are replaced with ordered k-tuples.
21
Exercise 1.7.10a We prove that the action is faithful for k < |A| or k ≥ |Z|. case 1) Suppose |A| is finite and k < |A|. We show that the action of Sn on k-element subsets of A is faithful. Let σx , σy ∈ Sn be any two distinct permutations of A. Since these permutations are distinct, there is some a ∈ A such that σx (a) 6= σy (a). From the invertibility of permutations this gives us the inequality σx (a) 6= σy (a) → σy−1 σx (a) 6= σy−1 (σy (a)) = a Since 1 ≥ k < n, we can choose a k-element subset B ⊂ A such that a ∈ B but σy−1 (σx (a)) 6∈ B. We can now demonstrate that the action is faithful, since σx (B) contains σx (a), but σy (B) does not contain σy (σy−1 (σx (a))) = σx (a). Thus σx and σy do not perform the same action on B. But these were arbitrary elements of SA , thus no two elements of SA perform the same action. By definition, this means that SA is faithful on the set of k-element subsets of A.
case 2) Suppose |A| is finite and k = |A|. There is only one distinct subset of size k: A itself. And every permutation of A is still the same set (just rearranged). Thus Sn is the trivial action on k-element subsets of |A|. case 3) Suppose |A| is infinite. For all finite k, we can follow the logic of case (1) and conclude that the action is faithful. If k is also infinite, then we can still choose an arbitrary a ∈ A and let B = A − {a}, and then follow the logic of case (1) and conclude that the action is still faithful.
Exercise 1.7.10b Choose two arbitrary permutations σx , σy ∈ Sn . For these to be distinct, there must be some a ∈ A such that σx (a) 6= σy (a). Having chosen such an a, let B be the k-tuple consisting of the element a repeated k times. It’s clear that σx (B) 6= σy (B), so the action of SA is faithful. But the value of k was never specified, so this proof holds for all values of k (finite and infinite).
Exercise 1.7.12 Note that a regular n-gon is a two-dimensional shape; the three-dimensional version is a regular n-hedron. Let n be even, and label the vertices of the n-gon clockwise as 0, 1, 2, . . . , n − 1. Let the n/2 pairs of opposite vertices be represented by the set of ordered pairs {ai } defined as n n no {ai } = i, + i : 0 ≤ i < 2 2 We define the action of D2n on the elements of {ai } so that the elements of D2n permute the set as follows: rk ai k
s ai
= a(i−k) mod n = a(i−kn/2) mod n
To prove this is an action, we note that r0 ai = ai for all ai , so property (i) of group actions is satisfied. To verify property (ii), let ra sb and rx sy be two arbitrary elements of D2n : (ra sb rx sy ) · ai = aj , with j = ((((i − ny/2) − x) − nb/2) − a) mod n
definition of ·, associativity of modular arithmetic
= ra sb · ak , with k = ((i − ny/2) − x) mod n
definition of ·, associativity of modular arithmetic
= ra sb · (rx sy · ai )
definition of · 0
The kernel of this action is {r } = {1}. 22
Exercise 1.7.13 We prove that the identity element of G is the only element in the kernel K. Property (i) of group actions guarantees that 1G ∈ K. But let g ∈ G be an arbitrary non-identity element of G and choose 1G ∈ A = G. Then g1G 6= 1G and so g 6∈ K.
Exercise 1.7.14 We prove by contradiction that property (ii) of group actions is not satisfied. Assume A = G is non-abelian and define the action g · a = ag. Let g1 , g2 be arbitrary elements of G. Hypothesis to be contradicted: suppose that property (ii) of group actions were satisfied. (g1 g2 ) · a = a(g1 g2 )
definition of this group action
= (ag1 )g2 )
associativity of operation on G
= g2 · (ag1 ))
definition of this group action
= g2 · (g1 · a))
definition of this group action
= (g2 g1 ) · a)) property (ii) of group actions This demonstrates that g1 commutes with g2 . But these were arbitrary elements of G so all elements of G commute, which means that G is abelian. This contradicts our initial assumption that property (ii) of group actions is satisfied.
Exercise 1.7.15 Let A = G and define the action g · a = ag −1 . The identity element 1G satisfies the condition 1 · a = a for all a ∈ A. To verify property (ii): (g1 g2 ) · a = a(g1 g2 )−1 = = =
a(g2−1 g1−1 ) (ag2−1 )g1−1 g1 · (ag2−1 )
= g1 · (g2 · a))
definition of this group action property of inverses associativity of operation on G definition of this group action definition of this group action
Exercise 1.7.16 Let A = G and define the action g · a = gag −1 . The identity element 1G satisfies the condition 1 · a = a for all a ∈ A. To verify property (ii): (g1 g2 ) · a = (g1 g2 )a(g1 g2 )−1 = = =
(g1 g2 )a(g2−1 g1−1 ) g1 (g2 (ag2−1 )g1−1 g1 · (g2 ag2−1 )
= g1 · (g2 · a))
definition of this group action property of inverses associativity of operation on G definition of this group action definition of this group action
Exercise 1.7.17 Let G be a group and fix a value for g ∈ G. Define a mapping f : G → G as f (x) 7→ gxg −1 . We want to prove that the given function is an isomorphism. We do so by showing that it’s a bijective homomorphism.
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f is a homomorphism f (xy) = g(xy)g −1 = g(xg −1 gy)g −1 = (gxg −1 )(gyg −1 ) = f (x)f (y) f is surjective Choose h ∈ G. Then g −1 hg ∈ G and f (g −1 hg) = h. Thus f is surjective f is injective: f (x) = 1 iff gxg −1 = 1 iff x = g −1 g = 1. The kernel of f consists only of the identity element; by exercise 1.6.14, this is sufficient to prove that f is injective.
Exercise 1.7.18 reflexive: 1 ∈ H and a = 1a so a ∼ a. transitive: If a ∼ b then a = hb, which implies b = h−1 a, which implies b ∼ a. symmetric: If a ∼ b and b ∼ c, then a = bh1 and b = ch2 , which implies a = ch2 h1 , which implies a ∼ c.
Exercise 1.7.19 Let H be a subgroup of finite group G, and let H act on G by left multiplication. Fix an element x ∈ G and let Ox be the orbit of x under the action of H. Define the map f : H → Ox as h 7→ hx. We prove that f is a bijection. f is surjective: Ox is defined as the set {hx : h ∈ G} which is exactly the image of f (H). f is injective: h(a) = h(b) iff ha = hb iff a = b (by left-cancellation). We’ve shown that |H| = |Ox |. But H was an arbitrary subgroup and x was an arbitrary element of G: thus |Ox | = |Oy | for all x, y ∈ G. Applying the preceeding exercise, we conclude that G can be partitioned into disjoint sets of size |Ox | = |H|; this can only occur if |G| is an integer multiple of |H|.
Exercise 1.7.20 Label the 4 vertices of the tetrahedron as 1, 2, 3, 4. Let G be the group of rigid motions of the tetrahedron. Each rigid motion corresponds to some permutation of the 4 vertices, so define the function ϕ : G → S4 so that ϕ(g) is the permutation corresponding to the rigid motion g. It’s clear that ϕ is a homomorphism: ϕ(g1 g2 ) = ϕ(g1 )ϕ(g2 ), since each side of the equation represents the rigid motion g1 followed by the rigid motion g2 . ϕ is injective: the kernel of ϕ consists only of the identity element of G, so ϕ is injective by exercise 1.6.14. Thus G is isomorphic to ϕ(G), which is a subgroup by exercise 1.6.13.
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Exercise 1.7.21 Choose one face of the cube to call the ’front’ and label its vertices a1 , a2 , a3 , a4 . On the opposite face, label the diagonally opposite points to be b1 , b2 , b3 , b4 . This labels all eight vertices of the cube. No matter how we rotate this cube, we know several things: • exactly four of these vertices will be on the front of the cube • for each i exactly one of ai , bi will be on the front of the cube (because they are diagonally opposite) • no rigid motion consists only of swapping one or more points with the points diagonally opposite (i.e., swap one or more ai with its corresponding bi ) This last point is particuarly important: it means that if there were a set of rigid motions that would give us front vertices of (arbitrary example) (a1 , b2 , b3 , a4 ), then none of the 15 other ordered 4-tuples we could form from replacing ai with bi (or vice-versa) would represent a rigid motion. This means that each rigid motion of the cube can be uniquely expressed as a permutation of (1, 2, 3, 4). We now follow the logic of the preceeding exercise to prove an isomorphism. Let G be the group of rigid motions of the cube. Each rigid motion corresponds to some permutation of the 4 pairs of opposite vertices, so define the function ϕ : G → S4 so that ϕ(g) is the permutation corresponding to the rigid motion g. It’s clear that ϕ is a homomorphism: ϕ(g1 g2 ) = ϕ(g1 )ϕ(g2 ), since each side of the equation represents the rigid motion g1 followed by the rigid motion g2 . ϕ is injective: the kernel of ϕ consists only of the identity element of G, so ϕ is injective by exercise 1.6.14. Thus G is isomorphic to ϕ(G), which is a subgroup by exercise 1.6.13.
Exercise 1.7.22 We can apply the preceeding exercise with opposite faces taking the place of opposite vertices. This is intuitively true: we could place a cube inside an octohedron by placing one vertex of the cube in the center of each face of the octohedron, and the rigid motions of the octohedron would correspond perfectly with the rigid motions of the cube. So the rigid motions of an octahedron are isomorphic to S4 itself (the errata for this textbook eliminates the ’subgroup’ qualification, although S4 is technically a subgroup of S4 ).
Exercise 1.7.23 Label the front, top, and one side face of the cube as (respectively) A, B, C. Label the respective opposite faces D, E, F . The set of rigid motions of the cube can be made up from various combinations of the three basic rotations: r1 = (A B D E), r2 = (A C D F ), r3 = (C B F E) When these rotations act on the set of opposite vertices, the action faithful. But when they act on the set of opposite faces, they are equivalent to r1 = ({A, D} {B, E}), r2 = ({A, D} {C, F }), r3 = ({C, F } {B, E}) so that r12 , r22 , and r32 are all equal to the identity motion. The kernel of the action becomes K = {i, r12 , r22 , r32 } (Any product of elements of K can be simplified to one of these elements). Note that 4|24, in concordance with exercise 19.
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