2.9 Factor Groups
1.
a. D6 = {1, a, a2, a3, a4, a5, b, ba, ba2, ba3, ba4, ba5}; K = Z(D6) = {1, a3} by Exercise 26 §2.6. The cosets are
We have 
, 
and KaKbKa = Kaba = Kb. Hence D6/K ≅ D3.
, and KaKbKa = Kaba = Kb. Hence D6/K ≅ D3.
c. G = A × B, K = {(a, 1) 
a 
A}. Note K G because (x, y)−1(a, 1)(x, y) = (x−1ax, 1) K. If b B, a A, note that (ab)(1, b)−1 K, so
a A}. Note K G because (x, y)−1(a, 1)(x, y) = (x−1ax, 1) K. If b B, a A, note that (ab)(1, b)−1 K, so
Thus G/K = {K(1, b) b B}. Observe that K(1, b)K(1, b′) = K(1, bb′) gives the Cayley table. Define θ : B → G/K by θ(b) = K(1, b). Then θ is onto, and it is 1 : 1 because θ(b) = θ(b1) gives K(1, b) = K(1, b1), so 
. Hence 
, b = b1. Finally, θ(b) · θ(b1) = K(1, b)K(1, b1) = K(1, bb1) = θ(bb1). So θ is an isomorphism; G/K ≅ B.
b B}. Observe that K(1, b)K(1, b′) = K(1, bb′) gives the Cayley table. Define θ : B → G/K by θ(b) = K(1, b). Then θ is onto, and it is 1 : 1 because θ(b) = θ(b1) gives K(1, b) = K(1, b1), so . Hence , b = b1. Finally, θ(b) · θ(b1) = K(1, b)K(1, b1) = K(1, bb1) = θ(bb1). So θ is an isomorphism; G/K ≅ B.
3. a. We have G =
a 
, so G/K = Ka . We claim 
. Certainly (Ka)12 = Ka12 = K. But 
, so, 
because Ka generates G/K. Now 
by Theorem 7 §2.4. Similarly, 
, and 
. Finally, Ka5 = (Ka)5 is a generator of G/K because gcd (5, 12) = 1 (Theorem 6 §2.4). Thus 
.
a , so G/K = Ka . We claim . Certainly (Ka)12 = Ka12 = K. But , so, because Ka generates G/K. Now by Theorem 7 §2.4. Similarly, , and . Finally, Ka5 = (Ka)5 is a generator of G/K because gcd (5, 12) = 1 (Theorem 6 §2.4). Thus .
4. a. G = a × b with o(a) = 8 and o(b) = 12, and K = (a2, b3) . We want 
in G/K. Then, as (a4, b)12 = (1, 1) K, 
, so 
. But since none of
a × b with o(a) = 8 and o(b) = 12, and K = (a2, b3) . We want in G/K. Then, as (a4, b)12 = (1, 1) K, , so . But since none of
or (a4, b)6 = (1, b6) are in K, 
.
.
5. a. We have Ka2 = K because a2 K. So o(Ka2) = 1. Next we have (Ka3)2 = Ka6 = K, so o(Ka3) divides 2. Since
K. So o(Ka2) = 1. Next we have (Ka3)2 = Ka6 = K, so o(Ka3) divides 2. Since
Similarly (Kba)2 = K(ba)2 = K and ba ∉ K, so o(Kba) = 2. Finally (Ka5)2 = Ka10 = K, so o(Ka5) = 2 because a5 ∉ K.
7. If 0 < n < m in 
, then 
because 
. Hence 
contains the infinite set 
. Now let 
be any element of 
. Then 
, so 
has finite order.
, then because . Hence contains the infinite set . Now let be any element of . Then , so has finite order.
9. If o(g) = n, then gn = 1, so (Kg)n = Kgn = K1 = K. Thus o(Kg) divides n.
11. Let o(g) = n, g G. Then (Kg)n = Kgn = K in G/K. On the other hand, 
, so (Kg)m = K by Lagrange's theorem. But gcd (m, n) = 1 implies 1 = nx + my, 
, so
G. Then (Kg)n = Kgn = K in G/K. On the other hand, , so (Kg)m = K by Lagrange's theorem. But gcd (m, n) = 1 implies 1 = nx + my, , so
Thus g K.
K.
13.
a. If z Z(G), then zK Z(G/K), so z K. Then z Z(K), so z = 1.
Z(G), then zK Z(G/K), so z K. Then z Z(K), so z = 1.
c. If g G, then 
for some n, so 
. Thus 
, so 
, so o(g) = pk for some k ≤ n + m.
G, then for some n, so . Thus , so , so o(g) = pk for some k ≤ n + m.
15. If Kg G/K, then g X , say 
, xi X, 
. Then Kg = (Kx1)k1(Kx2)k2
(Kxr)kr {Kx x X} .
G/K, then g X , say , xi X, . Then Kg = (Kx1)k1(Kx2)k2 (Kxr)kr {Kx x X} .
17.
a. If o(g) = n, o(g1) = m, then 
because G is abelian, and (g−1)n = 1.
because G is abelian, and (g−1)n = 1.
c. If G is torsion, it is clear that H and G/H are torsion. Conversely, if g G then (Hg)n = H for some n, so gn H. Thus (gn)m = 1 for some m ≠ 0, that is g T(G). Hence T(G) = G.
G then (Hg)n = H for some n, so gn H. Thus (gn)m = 1 for some m ≠ 0, that is g T(G). Hence T(G) = G.
19.
a. If G is abelian, then G/{1} is abelian, so G′ ⊆ {1}. Thus G′ = {1}.
c. Write
Let K = a2 = {1, a2, a4}. Then a2b = ba4 and a4b = ba2, so b−1Kb ⊆ K. Hence K D6 and D6/K is abelian (it has order 4). This means 
. But 
because D6 is not abelian. Thus 
.
a2 = {1, a2, a4}. Then a2b = ba4 and a4b = ba2, so b−1Kb ⊆ K. Hence K D6 and D6/K is abelian (it has order 4). This means . But because D6 is not abelian. Thus .
21. Given commutators [a, b] in G′ and [x, y] in H′, we have
If (g, h) G′ × H′, then (g, h) = (g, 1)(1, h) and each of (g, 1) and (1, h) are products of commutators of the form ([a, b], [x, y]). Hence G′ × H′ ⊆ (G × H)′. Conversely [(a, x), (b, y)] = ([a, x], [b, y]) shows that every commutator of G × H lies in G′ × H′, so (G × H)′ ⊆ G′ × H′.
G′ × H′, then (g, h) = (g, 1)(1, h) and each of (g, 1) and (1, h) are products of commutators of the form ([a, b], [x, y]). Hence G′ × H′ ⊆ (G × H)′. Conversely [(a, x), (b, y)] = ([a, x], [b, y]) shows that every commutator of G × H lies in G′ × H′, so (G × H)′ ⊆ G′ × H′.
23. a. The unity K = K1 of G/K is in H/K because 1 H. If Kh and Kh′ are in H/K where h, h′ H, then Kh · Kh′ = Khh′ H/K and also (Kh)−1 = Kh−1 H/K. Use the subgroup test.
H. If Kh and Kh′ are in H/K where h, h′ H, then Kh · Kh′ = Khh′ H/K and also (Kh)−1 = Kh−1 H/K. Use the subgroup test.
25. a. [Ka, Kb] = (Ka)(Kb)(Ka−1)(Kb−1) = K(aba−1b−1) = K[a, b].
27. Let σ : G → G be an automorphism. We must show σ(H) = H. Define 
by 
. Since K is characteristic in G, 
. Hence 
is well-defined and one-to-one; it is onto because σ is onto. Finally, 
. Hence σ is an automorphism of G/K, so 
by hypothesis. Hence, if h H, 
, h1 H, so 
. Thus σ(h) H, whence σ(H) ⊆ H. Similarly σ−1(H) ⊆ H and so H ⊆ σ(H). Thus H = σ(H).
by . Since K is characteristic in G, . Hence is well-defined and one-to-one; it is onto because σ is onto. Finally, . Hence σ is an automorphism of G/K, so by hypothesis. Hence, if h H, , h1 H, so . Thus σ(h) H, whence σ(H) ⊆ H. Similarly σ−1(H) ⊆ H and so H ⊆ σ(H). Thus H = σ(H).
1. a. Suppose 
is a prime. Then 
, so the group G/Z(G) is cyclic by Lagrange's theorem (Corollary 3). Then Theorem 2 shows G is abelian. Thus Z(G) = G, so 
, a contradiction.
is a prime. Then , so the group G/Z(G) is cyclic by Lagrange's theorem (Corollary 3). Then Theorem 2 shows G is abelian. Thus Z(G) = G, so , a contradiction.
29. Let n = kd and Dn = {1, a, . . ., an−1, b, ba, . . ., ban−1}, o(a) = n, o(b) = 2, aba = b. Let K = ak . Then K G because b−1akb = a−k K. In Dn/K, let 
, 
. Then 
, 
and
ak . Then K G because b−1akb = a−k K. In Dn/K, let , . Then , and
Since 
, Dn/K ≅ Dk.
, Dn/K ≅ Dk.
31. Let A ⊆ C4 and B ⊆ C ⊆ C8where |A| = |B| = 2 and |C| = 4 . (a) Let K = A × C and H = C4 × B. Then K ≅ C2 × C4 ≅ H, but 
while 
So 
while So
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