2.3 Subgroups
1.
a. No, 1 + 1 ∉ H.
c. No, 32 = 9 ∉ H.
e. No, (1 2)(3 4) · (1 3)(2 4) = (1 4)(2 3) ∉ H.
g. Yes, 0 = 6 
H. H is closed because it consists of the even residues in 
; −4 = 2, −2 = 4, so it is closed under inverses.
H. H is closed because it consists of the even residues in ; −4 = 2, −2 = 4, so it is closed under inverses.
i. Yes, the unity (0, 0) H. If (m, k) and (m′, k′) are in H, then so is (m, k) + (m′, k′) = (m + m′, k + k′) and −(m, k) = (− m, − k).
H. If (m, k) and (m′, k′) are in H, then so is (m, k) + (m′, k′) = (m + m′, k + k′) and −(m, k) = (− m, − k).
3. Yes. If H is a subgroup of G and K is a subgroup of H, then 1 K (it is the unity of H). If a, b K, then ab K because this is their product in H. Finally, a−1 is the inverse of a in H, hence in K.
K (it is the unity of H). If a, b K, then ab K because this is their product in H. Finally, a−1 is the inverse of a in H, hence in K.
5. a. We have 1 H because 1 = 12. If a, b H, then a−1 = a (because a2 = 1), so a−1 H. Finally, the fact that ab = ba gives (ab)2 = a2b2 = 1 · 1 = 1, so ab H.
H because 1 = 12. If a, b H, then a−1 = a (because a2 = 1), so a−1 H. Finally, the fact that ab = ba gives (ab)2 = a2b2 = 1 · 1 = 1, so ab H.
6.
a. We have 1 H because 1 = 12. If x, y H, write x = g2, y = h2. Then x−1 = (g−1)2 H and (since G is abelian) xy = g2h2 = (gh)2 H.
H because 1 = 12. If x, y H, write x = g2, y = h2. Then x−1 = (g−1)2 H and (since G is abelian) xy = g2h2 = (gh)2 H.
c. The set of squares in A4 consists of ε and all the 3-cycles. This is not a subgroup since (1 2 3)(1 2 4) = (1 3)(2 4).
7. a. We have 1 = g0 
g 
. If x and y are in g, write x = gk, y = gm, 
. Then x−1 = g−k g and xy = gk+m g . Use the subgroup test.
g . If x and y are in g, write x = gk, y = gm, . Then x−1 = g−k g and xy = gk+m g . Use the subgroup test.
8. a. If x X (since X is nonempty), 1 = xx−1 X . Clearly X is closed, and if 
, then 
. Hence X is a subgroup; clearly X⊆ X .
X (since X is nonempty), 1 = xx−1 X . Clearly X is closed, and if , then . Hence X is a subgroup; clearly X⊆ X .
9. We have 1 C(g) because 1g = g1. If 
, then
C(g) because 1g = g1. If , then
This shows zw C(g). Finally zg = gz implies g = z−1gz, so gz−1 = z−1g. Thus z−1 C(g). Use the subgroup test.
C(g). Finally zg = gz implies g = z−1gz, so gz−1 = z−1g. Thus z−1 C(g). Use the subgroup test.
11. We have 
. If X, Y G, write 
, 
. Then 
and 
. Use Theorem 1.
. If X, Y G, write , . Then and . Use Theorem 1.
13. a. Clearly the unity (1, 1) of G × G is in H. If x, y H, write x = (f, f) and y = (g, g). Then xy = (fg, fg) H and x−1 = (f−1, f−1) H. So H is a subgroup of G by Theorem 1.
H, write x = (f, f) and y = (g, g). Then xy = (fg, fg) H and x−1 = (f−1, f−1) H. So H is a subgroup of G by Theorem 1.
15.
a. C5 = {1, g, g2, g3, g4}, g5 = 1. If H ≠ {1} is a subgroup, one of g, g2, g3, g4 is in H. If g H, then H = C5. But g = (g2)3 = (g3)2 = (g4)4, so H = C5 in any case. Hence {1} and C5 are the only subgroups.
H, then H = C5. But g = (g2)3 = (g3)2 = (g4)4, so H = C5 in any case. Hence {1} and C5 are the only subgroups.
c. S3 = {1, σ, σ2, τ, τσ, τσ2}, σ3 = 1 = τ2, στ = τσ2. We claim {1}, {1, σ, σ2}, {1, τ}, {1, τσ} and {1, τσ2} are all the proper subgroups. They are subgroups by Theorem 2. Suppose a subgroup H is not one of these: Case 1.σ H or σ2 H. Then {1, σ, σ2} ⊆ H, so H contains one of τ, τ, σ, τσ2. But τ = (τσ)σ2 = τ(τσ2)σ, so σ H and τ H. This means H = S3. Case 2.σ6 H and σ26 H. Then H contains two of τ, τσ, τσ2. But τ(τσ) = σ, τ(τσ2) = σ2 and (τσ)(τσ2) = τ(τσ2)σ2 = σ, so this case cannot occur.
H or σ2 H. Then {1, σ, σ2} ⊆ H, so H contains one of τ, τ, σ, τσ2. But τ = (τσ)σ2 = τ(τσ2)σ, so σ H and τ H. This means H = S3. Case 2.σ6 H and σ26 H. Then H contains two of τ, τσ, τσ2. But τ(τσ) = σ, τ(τσ2) = σ2 and (τσ)(τσ2) = τ(τσ2)σ2 = σ, so this case cannot occur.
16. a.1 H ∩ K because 1 H and 1 K. If a H ∩ K, then a H and a K. Thus a−1 H and a−1 K, so a−1 H ∩ K. If b H ∩ K also, then b H, and b K, so ab H and ab K. Thus ab H ∩ K.
H ∩ K because 1 H and 1 K. If a H ∩ K, then a H and a K. Thus a−1 H and a−1 K, so a−1 H ∩ K. If b H ∩ K also, then b H, and b K, so ab H and ab K. Thus ab H ∩ K.
17. If H ⊆ K or K ⊆ H, then H ∪ K is K or H respectively, so H ∪ K is a subgroup. Conversely, suppose H ∪ K is a subgroup and H 
K. We show K ⊆ H. If k K, we must show that k H. Choose h H K. Then kh ∉ K (if kh = k1 K, then h = k−1k1 K). Since kh is in H ∪ K (because H ∪ K is a subgroup), this gives kh H. But kh = h1 implies k = h−1h1 H, as required.
K. We show K ⊆ H. If k K, we must show that k H. Choose h H K. Then kh ∉ K (if kh = k1 K, then h = k−1k1 K). Since kh is in H ∪ K (because H ∪ K is a subgroup), this gives kh H. But kh = h1 implies k = h−1h1 H, as required.
19. a. g−1Hg = g−1gH = 1H = H. So the only conjugate of H is H itself.
20. a. If g G, then gh = hg for all h H (since H ⊆ Z(G)), so gH = Hg. Hence g−1Hg = g−1gH = 1H = H.
G, then gh = hg for all h H (since H ⊆ Z(G)), so gH = Hg. Hence g−1Hg = g−1gH = 1H = H.
21. Let 
. Then 
for all a, b, c. This means ay + bz = xb + yc for all a, b, c. If we take a = c = 0 and b = 1, we get x = z. Take a = 1and b = c = 0 to get y = 0. Hence
. Then for all a, b, c. This means ay + bz = xb + yc for all a, b, c. If we take a = c = 0 and b = 1, we get x = z. Take a = 1and b = c = 0 to get y = 0. Hence
22. Let 
be in 
. Then ZA = AZ for all 
. Taking 
leads to y = z and 
; taking 
leads to z = 0 (and 
). Hence 
. Each matrix 
is central because (xI)A = xA = A(xI) for all A.
be in . Then ZA = AZ for all . Taking leads to y = z and ; taking leads to z = 0 (and ). Hence . Each matrix is central because (xI)A = xA = A(xI) for all A.
23. Yes. If σ = (1 2 3), then H = {ε, σ, σ2} is an abelian subgroup of S3, but Z(S3) = {ε}.
25. Assume that KH ⊆ HK. Then 1 = 1 1 HK. If h H and k K, then (hk)−1 = k−1h−1 KH ⊆ HK, and
HK. If h H and k K, then (hk)−1 = k−1h−1 KH ⊆ HK, and
Conversely, if HK is a subgroup then kh = (h−1k−1)−1 HK.
HK.
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.