8.1 Products and Factors
1. a. XY = {τ, τσ}{τ, τσ2} = {τ2, τ2σ2, τστ, τστσ2} = {ε, σ2, τ2σ2, τ2σ4} = {ε, σ2, σ} . YX = {τ, τσ2}{τ, τσ} = {τ2, τ2σ, τσ2τ, τσ2τσ} = {ε, σ, τ2σ, τ2σ2} = {ε, σ, σ2}.
3. If G′ ⊆ H, then H/G′ 
G/G′ because G/G′ is abelian. Hence H G by the correspondence theorem.
G/G′ because G/G′ is abelian. Hence H G by the correspondence theorem.
4.
a. G = D6 = {1, a, . . ., a5, b, ba, . . ., ba5}, o(a) = 6, o(b) = 2, aba = b. We have K = Z(D6) = {1, a3}. Write 
, 
. Then 
, 109 
, 
, 
. Hence G/K ≅ D3 and the only subgroups of G/K are 
and G/K. So the subgroups of G containing K are G, K and
, . Then , 109 , , . Hence G/K ≅ D3 and the only subgroups of G/K are and G/K. So the subgroups of G containing K are G, K and
Note that H1 ≅ H2 ≅ H3 ≅ K4 —the Klein group (in contrast with (b)).
c. G = A4 and K = {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. Then 
, so 
is cyclic of order 3. Hence the only subgroups of A4 containing K are K and A4. In particular, K is maximal normal in A4.
, so is cyclic of order 3. Hence the only subgroups of A4 containing K are K and A4. In particular, K is maximal normal in A4.
5.
a. Every subgroup of 
has the form 
, 
, and 
. This is simple (abelian) if and only if n is a prime. Thus 
a prime} are the maximal normal subgroups of 
.
has the form , , and . This is simple (abelian) if and only if n is a prime. Thus a prime} are the maximal normal subgroups of .
c. G = D10 = {1, a, . . ., a9, b, ba, . . ., ba9}, o(a) = 10, o(b) = 2, aba = b. If H is maximal normal in G, then G/H is abelian (it's order is ≤5) and so has order c, a prime dividing 20. Thus 
so 
. The subgroups of order 10 are
so . The subgroups of order 10 are
There is no element of order 4, so the subgroups of order 4 are
7. If K G, G/K cyclic, 
, 
, let k 
m and m n. Then 
and 
divides 
. Hence G/K cyclic implies it has a unique subgroup X with 
. Write X = H/K. Then 
, so 
. If K ⊆ H1 ⊆ G and 
, then 
, so H′/K = H/K by the uniqueness. Thus H1 = H.
G, G/K cyclic, , , let k m and m n. Then and divides . Hence G/K cyclic implies it has a unique subgroup X with . Write X = H/K. Then , so . If K ⊆ H1 ⊆ G and , then , so H′/K = H/K by the uniqueness. Thus H1 = H.
9. We must prove that (H1/K) ∩ (H2/K) = (H1 ∩ H2)/K. We have
by Lemma 2. If Kg 
(H1/K) ∩ (H2/K), let Kg = Khi for hi Hi, i = 1, 2 . Then g Khi ⊆ Hi for each i, so g H1 ∩ H2. This shows that Kg (H1 ∩ H2)/K, and so proves that
(H1/K) ∩ (H2/K), let Kg = Khi for hi Hi, i = 1, 2 . Then g Khi ⊆ Hi for each i, so g H1 ∩ H2. This shows that Kg (H1 ∩ H2)/K, and so proves that
10. a. If HH1 = H1H, then 
. Hence 
, and the other inclusion is similar. Conversely, if 
then 
; so 
because KH1 ⊆ H1. Thus HH1 ⊆ H1H and the other inclusion is similar.
. Hence , and the other inclusion is similar. Conversely, if then ; so because KH1 ⊆ H1. Thus HH1 ⊆ H1H and the other inclusion is similar.
11. 
X∪ Y 
is a subgroup containing both X and Y, so X ⊆ X ∪ Y and Y ⊆ X ∪ Y . Then X Y ⊆ X ∪ Y because X∪ Y is closed. If X Y = X ∪ Y , then X Y is a subgroup, so X Y = Y X Lemma 2 §2.8. Conversely, if X Y = Y X then X Y is a subgroup (again by Lemma 2 §2.8). Since X Y contains both X and Y, it contains X ∪ Y, and so X∪ Y ⊆ X Y by Theorem 8 §2.4.
X∪ Y is a subgroup containing both X and Y, so X ⊆ X ∪ Y and Y ⊆ X ∪ Y . Then X Y ⊆ X ∪ Y because X∪ Y is closed. If X Y = X ∪ Y , then X Y is a subgroup, so X Y = Y X Lemma 2 §2.8. Conversely, if X Y = Y X then X Y is a subgroup (again by Lemma 2 §2.8). Since X Y contains both X and Y, it contains X ∪ Y, and so X∪ Y ⊆ X Y by Theorem 8 §2.4.
12. a. H2 ⊆ H because H is closed; H ⊆ H2 because 1 H.
H.
13. Write 
Since H ∩ K is a subgroup of H we have m {1, p, q, pq} by Lagrange's theorem. Similarly m {1, q, r, qr}, so m {1, q} . But if m = 1 then 
a contradiction. So m = q as required.
Since H ∩ K is a subgroup of H we have m {1, p, q, pq} by Lagrange's theorem. Similarly m {1, q, r, qr}, so m {1, q} . But if m = 1 then a contradiction. So m = q as required.
15. Note that KA and KB are subgroups by Theorem 5 §2.8, and KA = AK, KB = BK. If kb KB, we have Ab = bA and Kb = bK. Hence
KB, we have Ab = bA and Kb = bK. Hence
Thus KA KB.
KB.
17. Given K G, 
, assume H is any subgroup of G with 
. Then HK is a subgroup and HKK ≅ HH ∩ K by the second isomorphism theorem. Since 
, we have 
for some k such that 0 ≤ k ≤ n. Hence 
. But HKK is a subgroup of GK, and 
. It follows that pk divides m Since p
m by hypothesis, we have k = 0. Thus H = H ∩ K and HK = K; so certainly K ⊆ H. Hence H = K because 
G, , assume H is any subgroup of G with . Then HK is a subgroup and HKK ≅ HH ∩ K by the second isomorphism theorem. Since , we have for some k such that 0 ≤ k ≤ n. Hence . But HKK is a subgroup of GK, and . It follows that pk divides m Since pm by hypothesis, we have k = 0. Thus H = H ∩ K and HK = K; so certainly K ⊆ H. Hence H = K because
18.
a. We have M ⊂ KM because M = KM implies K ⊆ M. Since KM is normal, we have M = KM because M is maximal normal.
c. By (a) and Theorem 3, 
so (c) follows from (b).
so (c) follows from (b).
19. a. The argument in Example 3 goes through with “cyclic” replaced by “abelian”. All that is needed is that subgroups and factor groups of abelian groups are again abelian. However, we give a slick argument using (b) below. Let G′ be metabelian, that is G′ is abelian by (b). If H is a subgroup of G, then H′ ⊆ G′ (since commutators in H are commutators in G), so H′ is abelian. Thus H is metabelian. Now suppose that N G. Then in GN, each commutator [Na, Nb] = N[a, b] NG′. Hence 
Since 
is abelian (being a factor group of the abelian group G′), it follows that 
is abelian, that is 
is metabelian.
G. Then in GN, each commutator [Na, Nb] = N[a, b] NG′. Hence Since is abelian (being a factor group of the abelian group G′), it follows that is abelian, that is is metabelian.
21. We are given subgroups H and K where 
, 
, p ≠ q primes. Since H ∩ K ⊆ K and 
, we have 
, q or q2 by Lagrange's theorem. Similarly H ∩ K ⊆ H and 
shows that 
p, q or pq. Hence 
or q, and it remains to show 
. But 
gives (by Theorem 4) 
. Since HK ⊆ G, this contradicts 
.
, , p ≠ q primes. Since H ∩ K ⊆ K and , we have , q or q2 by Lagrange's theorem. Similarly H ∩ K ⊆ H and shows that p, q or pq. Hence or q, and it remains to show . But gives (by Theorem 4) . Since HK ⊆ G, this contradicts .
22. a. Let a ≠ b be elements of order 2. Since G is abelian, H = {1, a, b, ab} is closed, and so is a subgroup. Then 
divides 
by Lagrange's theorem.
divides by Lagrange's theorem.
23. The unity of M is {1} = 1. So if X is a unit in M, let XY = 1. This means that xy = 1 for all x X, y Y. So, given y Y, x = y−1 for all x X. Thus X is a singleton, say X = {g} = g G. Conversely, each g G is a unit in M because gg−1 = 1.
X, y Y. So, given y Y, x = y−1 for all x X. Thus X is a singleton, say X = {g} = g G. Conversely, each g G is a unit in M because gg−1 = 1.
24.
a. We must show 
. If σ A = aut G and 
(τa(g) = ag for all g G), then στa(g) = σ(ag) = σ(a) · σ(g) = τσ(a)σ(g). Thus στa = τσ(a)σ, so 
. If we take b = σ(a), this is 
, so 
.
. If σ A = aut G and (τa(g) = ag for all g G), then στa(g) = σ(ag) = σ(a) · σ(g) = τσ(a)σ(g). Thus στa = τσ(a)σ, so . If we take b = σ(a), this is , so .
c. Let 
and 
, say λ = τbσ. Then (using τaσ = στσ(a) in (a)):
and , say λ = τbσ. Then (using τaσ = στσ(a) in (a)):
Thus 
.
.
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