8.3 Group Actions
1. a. If |G| = 20, there is a 
G with o(a) = 5 by Cauchy's theorem. Thus |G : 
a 
|= 4 so there is a homomorphism θ : G → S4 with ker θ ⊆ a . Hence | a /ker θ| divides 24 so ker θ ≠ {1}. Thus ker θ = a because o(a) is prime, so a 
G.
G with o(a) = 5 by Cauchy's theorem. Thus |G : a |= 4 so there is a homomorphism θ : G → S4 with ker θ ⊆ a . Hence | a /ker θ| divides 24 so ker θ ≠ {1}. Thus ker θ = a because o(a) is prime, so a G.
3. Assume p ≤ q. By Cauchy's theorem, let a G, o(a) = q. Then H = a has index p so H G by the Corollary to Theorem 1.
G, o(a) = q. Then H = a has index p so H G by the Corollary to Theorem 1.
5. If |H| = p then |G : H| = m so let θ : G → Sm be a homomorphism with ker θ ⊆ H. Now ker θ = {1} is impossible since |G| = pm does not divide m !. Since |H| = p, the only other possibility is ker θ = H, so H G as asserted.
G as asserted.
6. a. If |An : H| = p let θ : An → Sp be a homomorphism with ker θ ⊆ H. Then θ is one-to-one because An is simple, so 
divides p !, say 
. But 
so 
and hence p ≤ n. By hypothesis p < n. Hence 
so q = 1, n ! = p !, a contradiction since n ≥ 5.
divides p !, say . But so and hence p ≤ n. By hypothesis p < n. Hence so q = 1, n ! = p !, a contradiction since n ≥ 5.
7. If U ⊆ V are subgroups then g−1Ug ⊆ g−1Vg for all g G, so core U⊆ core V. Hence core(H∩ K) ⊆ core H ∩ score K. If x core H∩ core K, let g G. Thus g−1(H ∩ K)g = (g−1Hg) ∩ (g−1Kg) and so x g−1(H ∩ K)g. Thus x core (H ∩ K).
G, so core U⊆ core V. Hence core(H∩ K) ⊆ core H ∩ score K. If x core H∩ core K, let g G. Thus g−1(H ∩ K)g = (g−1Hg) ∩ (g−1Kg) and so x g−1(H ∩ K)g. Thus x core (H ∩ K).
9. This is Exercise 26 §2.8.
10. a. H0 ⊆ H because H = 1G(H). If τ autG then τ−1σ autG for all σ autG, so H0 ⊆ τ−1σ(H). Thus τ(H0) ⊆ σ(H) for all σ, so τ(H0) ⊆ H0. Similarly τ−1(H0) ⊆ H0, whence τ(H0) = H0. Thus H0 is characteristic in G.
autG then τ−1σ autG for all σ autG, so H0 ⊆ τ−1σ(H). Thus τ(H0) ⊆ σ(H) for all σ, so τ(H0) ⊆ H0. Similarly τ−1(H0) ⊆ H0, whence τ(H0) = H0. Thus H0 is characteristic in G.
11. (1) ⇒ (2). If X is a nontrivial finite G-set, let θ : G → SX be the homomorphism in Theorem 2. Then θ(G) ≠ {1X} because the action is not trivial, so ker θ ≠ G. Clearly 
is finite.
is finite.
13. We have 0 · z = ei0z = z and b · (a · z) = eib(eiaz) = ei(a+b)z = (a + b) · z, so it is indeed an action. If z = reiθ then a · z = rei(θ+a), so the action by a is to rotate z about the origin counterclockwise through a radians. If 
, the orbit G · z = {a · z 
a G} is the circle, center at 0, radius |z|. Given 
, the stabilizer is S(z) = {a a · z = z} = {a zeia = z}. Hence S(0) = G; and if z ≠ 0, then 
, the orbit G · z = {a · z a G} is the circle, center at 0, radius |z|. Given , the stabilizer is S(z) = {a a · z = z} = {a zeia = z}. Hence S(0) = G; and if z ≠ 0, then
15. If σ = (k1k2
)(m1m2 )(n1n2 ) , the orbits are G · k1 = {k1, k2, . . . }, G · m1 = {m1, m2, . . . }, G · n1 = {n1, n2, . . . } and so on. Clearly G · k = {k} if and only if σ fixes k.
)(m1m2 )(n1n2 ) , the orbits are G · k1 = {k1, k2, . . . }, G · m1 = {m1, m2, . . . }, G · n1 = {n1, n2, . . . } and so on. Clearly G · k = {k} if and only if σ fixes k.
17. We have x ≡ x because x = 1 · x; if x ≡ y then y = a · x for some a G, so x = a−1 · y, y ≡ x; if x ≡ y and y ≡ z then y = a · x, z = b · y, so z = b · (a · x) = (ba) · x, that is x ≡ z.
G, so x = a−1 · y, y ≡ x; if x ≡ y and y ≡ z then y = a · x, z = b · y, so z = b · (a · x) = (ba) · x, that is x ≡ z.
19. Given G and X = {H H is a subgroup of G}, define a · H = aHa−1. Since aHa−1 is again in X, this is an action. The fixer here is {a G a · H = H for all H} = {a G aHa−1 = H for all H}. This contains Z(G). If G = Q is the quaternion group, then every subgroup is normal, so F = Q. However Z(G) = {1, − 1}.
H is a subgroup of G}, define a · H = aHa−1. Since aHa−1 is again in X, this is an action. The fixer here is {a G a · H = H for all H} = {a G aHa−1 = H for all H}. This contains Z(G). If G = Q is the quaternion group, then every subgroup is normal, so F = Q. However Z(G) = {1, − 1}.
21. Let X = {xH x G} and let g · xH = gxH. Then H X and
x G} and let g · xH = gxH. Then H X and
23.
a. If a, b S(x) then (ab) · x = a · (b · x) = a · x = x and,
S(x) then (ab) · x = a · (b · x) = a · x = x and,
so ab, a−1 S(G). Since 1 · x = x, 1 S(x) and we are done.
S(G). Since 1 · x = x, 1 S(x) and we are done.
c. Assume bS(y)b−1 = S(x). Define σ : G · x → G · y by σ(g · x) = bg · y. Then
Thus σ is well defined and one-to-one; it is clearly onto. Note that if G is finite this can be proved as follows:
24. a. The fixer is F = ∩ xXS(x). If y X then G · x = X = G · y by hypothesis, so S(x) and S(y) are conjugate subgroups by the preceding exercise, say S(y) = bS(x)b−1. Then K ⊆ S(x) gives
XS(x). If y X then G · x = X = G · y by hypothesis, so S(x) and S(y) are conjugate subgroups by the preceding exercise, say S(y) = bS(x)b−1. Then K ⊆ S(x) gives
Thus K ⊆ ∩ yXS(y) = F.
XS(y) = F.
25. This action is well defined: If a · x = b · x then
It is clearly an action.
27. Let G act on X = {H H ⊆ G is a subgroup} by conjugation: a · H = aHa−1. Then Xf = {H a · H = H for all a} is the set of normal subgroups of G. Hence the nonnormal subgroups are partitioned into nonsingleton conjugacy classes, and each of these has |G : N(H)| elements for some H. Since G is a p-group, this is a multiple of p, and the result follows.
H ⊆ G is a subgroup} by conjugation: a · H = aHa−1. Then Xf = {H a · H = H for all a} is the set of normal subgroups of G. Hence the nonnormal subgroups are partitioned into nonsingleton conjugacy classes, and each of these has |G : N(H)| elements for some H. Since G is a p-group, this is a multiple of p, and the result follows.
29. If |G : H| = p then |G : σ−1(H)| = p for all σ autG. Hence K ⊆ σ−1(H) so σ(K) ⊆ H. It follows that σ(K) ⊆ K, so K is characteristic; in particular normal. Now if G is a p-group, Hi ≠ N(Hi) for each i by Theorem 5, so since |G : Hi| = p is prime, Hi G by Exercise 6 §8.2 and |G/Hi| = p. Let A = G/H1 × × G/Hm and define θ : G → A by θ(g) = (gH1, . . ., gHm). This is a homomorphism with ker θ = K, so G/K ≅ θ(G) ⊆ A. The result follows.
autG. Hence K ⊆ σ−1(H) so σ(K) ⊆ H. It follows that σ(K) ⊆ K, so K is characteristic; in particular normal. Now if G is a p-group, Hi ≠ N(Hi) for each i by Theorem 5, so since |G : Hi| = p is prime, Hi G by Exercise 6 §8.2 and |G/Hi| = p. Let A = G/H1 × × G/Hm and define θ : G → A by θ(g) = (gH1, . . ., gHm). This is a homomorphism with ker θ = K, so G/K ≅ θ(G) ⊆ A. The result follows.
31. We have 1 · (h, k) = (h, k), and
Thus it is an action. Let X = H × K so |X| = |H| |K|. Let A = H ∩ K and define λ : A → A · (h, k) by λ(a) = a · (h, k) = (ha−1, ak). This is clearly a bijection, so every orbit has |A| elements. Finally, define
by (hk)μ = A · (h, k). Then
so μ is well-defined. Conversely, a · (h, k) = (h1, k1) implies hk = h1k1, so μ is a bijection.
32.
a. (1, 1) · x = 1x1−1 = x, 
= (h1h, k1k) · x = [(h1k1) · (h, k)] · x. The orbit is (H × K) · x = {(h, k) · x h H, k K} = {hxk−1 h H, k K} = HxK—a double coset.
= (h1h, k1k) · x = [(h1k1) · (h, k)] · x. The orbit is (H × K) · x = {(h, k) · x h H, k K} = {hxk−1 h H, k K} = HxK—a double coset.
c. The size of the double coset is
by (b). Frobenius' theorem follows.
33. a. Any two cosets aH and bH are in the same orbit because bH = (ba−1) · (aH).
34. a. This action is well-defined because [x] = [x1] gives ϕ(x) = ϕ(x1) so, since ϕ is a G-morphism,
Thus [a · x] = [a · x1] so the action is well-defined. Now 1 · [x] = [1 · x] = [x] and a · (b · [x]) = a · [b · x] = [a · (b · x)] = [(ab) · x] = (ab) · [x]. Hence 
is a G-set.
is a G-set.
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