2.6 Cosets and Lagrange's Theorem
1.
a. 1H = H1 = {1, a4, a8, a12, a16} aH = Ha = {a, a5, a9, a13, a17} a2H = Ha2 = {a2, a6, a10, a14, a18} a3H = Ha3 = {a3, a7, a11, a15, a19} 1K = K1 = {1, a2, a4, a6, a8, a10, a12, a14, a16, a18} aK = Ka = {a, a3, a5, a7, a9, a11, a13, a15, a17, a19}
c. 
e. 
3. No. If H = {1, b} ⊆ D3, then Ha = {a, ba) = Hba, but aH = {a, ba2} ≠ baH.
5. a. a ≡ a because a−1a = 1 
H. If a ≡ b then b−1a H, whence a−1b = (b−1a)−1 H, so b ≡ a. Finally, if a ≡ b and b ≡ c then b−1a H and c−1b H, so c−1a = (c−1b)(b−1a) H. Thus a ≡ c.
H. If a ≡ b then b−1a H, whence a−1b = (b−1a)−1 H, so b ≡ a. Finally, if a ≡ b and b ≡ c then b−1a H and c−1b H, so c−1a = (c−1b)(b−1a) H. Thus a ≡ c.
7. If Ha = bH, then a Ha gives a bH, so aH = bH. Thus aH = Ha. Similarly bH = Hb, so aH = Ha = bH = Hb.
Ha gives a bH, so aH = bH. Thus aH = Ha. Similarly bH = Hb, so aH = Ha = bH = Hb.
9.
a. If 
, then 
equals 
or 
, according as x > 0 or x < 0. Here 
is the set of positive real numbers, and 
is the set of negative real numbers.
, then equals or , according as x > 0 or x < 0. Here is the set of positive real numbers, and is the set of negative real numbers.
c. If 
, write x = n + t, 
, 0 ≤ t < 1. Then 
consists of all points on the line at distance t to the right of an integer.
, write x = n + t, , 0 ≤ t < 1. Then consists of all points on the line at distance t to the right of an integer.
10. a. Write H =
a6 
. Then 
, so 
.
a6 . Then , so .
11. a. (H ∩ K)a ⊆ Ha ∩ Ka is clear. If x Ha ∩ Ka, write x = ha = ka, h H, k K. Then h = k by cancellation, so h H ∩ K. Thus x = ha (H ∩ K)a.
Ha ∩ Ka, write x = ha = ka, h H, k K. Then h = k by cancellation, so h H ∩ K. Thus x = ha (H ∩ K)a.
12.
a. If o(g) = m, we show m = 12. We have m 
12 by Lagrange's theorem, so m is one of 1, 2, 3, 4, 6 or 12. If m ≠ 12, then m|4 or m|6, so g4 = 1 or g6 = 1, contrary to hypothesis.
12 by Lagrange's theorem, so m is one of 1, 2, 3, 4, 6 or 12. If m ≠ 12, then m|4 or m|6, so g4 = 1 or g6 = 1, contrary to hypothesis.
c. Now o(g) divides 60, so is one of 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Again each of these except 60 divides one of 12, 20 or 30, so o(g) = 60.
13. If 
, then m|12 and 4|m (because K ⊆ H). Thus 
or 
; that is H = K or H = A4.
, then m|12 and 4|m (because K ⊆ H). Thus or ; that is H = K or H = A4.
5. Since H ∩ K ⊆ H, we have that 
divides the prime 
. So either 
(and H ∩ K = {1}) or 
(so H ∩ K = H, that is H ⊆ K).
divides the prime . So either (and H ∩ K = {1}) or (so H ∩ K = H, that is H ⊆ K).
1. a. We have gn = 1 by Lagrange's theorem. Since gcd (m, n) = 1, write 1 = xn + ym; 
. If gm = 1, then g = g1 = (gm)x(gn)y = 1x1y = 1.
. If gm = 1, then g = g1 = (gm)x(gn)y = 1x1y = 1.
17. Let H be a subgroup of G, H ≠ G. Then 
or p by Lagrange's Theorem. If 
, then H = 1 . If 
then H is cyclic by Corollary 3 of Lagrange's Theorem.
or p by Lagrange's Theorem. If , then H = 1 . If then H is cyclic by Corollary 3 of Lagrange's Theorem.
19. If x = ak = bk, then x a ∩ b . But a ∩ b = {1} by Corollary 4 of Lagrange's Theorem because 
and 
are relatively prime. Thus ak = 1 and bk = 1, so m k and n k. Thus mn k, again because m and n are relatively prime.
a ∩ b . But a ∩ b = {1} by Corollary 4 of Lagrange's Theorem because and are relatively prime. Thus ak = 1 and bk = 1, so m k and n k. Thus mn k, again because m and n are relatively prime.
21. If n = 1 then 
. If n ≥ 2 and 
, then k = qn + r, 0 ≤ r ≤ n − 1, so 
. Thus 
; that is the cosets are
. If n ≥ 2 and , then k = qn + r, 0 ≤ r ≤ n − 1, so . Thus ; that is the cosets are
These are distinct. For if 
with 0 ≤ r ≤ s ≤ n − 1 then 
, so 0 ≤ s − r ≤ s ≤ n − 1. But s − r ≥ 0 is a multiple of n; so s − r = 0. Thus there are exactly n cosets.
with 0 ≤ r ≤ s ≤ n − 1 then , so 0 ≤ s − r ≤ s ≤ n − 1. But s − r ≥ 0 is a multiple of n; so s − r = 0. Thus there are exactly n cosets.
23. Let g G, g ≠ 1. Then o(g) divides 
, say o(g) = pm, m ≤ k. Since m ≠ 0, 
by Theorem 5 §2.4.
G, g ≠ 1. Then o(g) divides , say o(g) = pm, m ≤ k. Since m ≠ 0, by Theorem 5 §2.4.
25. a. For k ≥ 1 we induct on k. It is given for k = 1 . If akbak = b for some k ≥ 0, then ak+1bak+1 = aba = a. Thus akbak = b for all k ≥ 1 by induction. It is clear if k = 0 . But a−1ba−1 = a−1(aba)a−1 = b, and a similar argument shows a−kba−k = b if k ≥ 1.
27. No. D5 × C3 has no element of order 10, while D3 × C5 has 12 elements of order 10.
29. a. H ∩ K is a subgroup of H so H ∩ K = {1} or H ∩ K = H by Lagrange's Theorem. Similarly, H ∩ K ⊆ K implies that H ∩ K = {1} or H ∩ K = K. Thus H ∩ K ≠ {1} implies H = H ∩ K = K.
31. If 
, let Kh1, . . ., Khn be the distinct cosets of K in H. Thus H = Kh1 ∪ 
∪ Khn, a disjoint union. Then Hg ⊆ Kh1g ∪ ∪Khng is clear, and it is equality because K ⊆ H. Thus each H-coset in G is the union of nK-cosets. If 
this gives 
. Conversely, if 
is finite, then 
is clearly finite and 
is finite by the hint since each H-coset is a union of K -cosets.
, let Kh1, . . ., Khn be the distinct cosets of K in H. Thus H = Kh1 ∪ ∪ Khn, a disjoint union. Then Hg ⊆ Kh1g ∪ ∪Khng is clear, and it is equality because K ⊆ H. Thus each H-coset in G is the union of nK-cosets. If this gives . Conversely, if is finite, then is clearly finite and is finite by the hint since each H-coset is a union of K -cosets.
32. a. (H ∩ K)g ⊆ Hg ∩ Kg is clear. If x Hg ∩ Kg, write x = hg = kg, h H, k K. Then h = k H ∩ K by cancellation, so x = hg (H ∩ K)g. Hence (H ∩ K)g = Hg ∩ Kg, so each (H ∩ K)g coset is the intersection of one of the mH-cosets with one of the nK-cosets. There are thus at most mnH ∩ K-cosets.
Hg ∩ Kg, write x = hg = kg, h H, k K. Then h = k H ∩ K by cancellation, so x = hg (H ∩ K)g. Hence (H ∩ K)g = Hg ∩ Kg, so each (H ∩ K)g coset is the intersection of one of the mH-cosets with one of the nK-cosets. There are thus at most mnH ∩ K-cosets.
33. If H1, , Hn are all of finite index in G, we show H1 ∩ ∩ Hk is of finite index for each k = 1, 2, . . ., n. This is clear if k = 1. If it holds for some k, then H1 ∩ ∩ Hk+1 = (H1 ∩ ∩ Hk) ∩ Hk+1 is a finite index by part (a) of the preceding exercise.
, Hn are all of finite index in G, we show H1 ∩ ∩ Hk is of finite index for each k = 1, 2, . . ., n. This is clear if k = 1. If it holds for some k, then H1 ∩ ∩ Hk+1 = (H1 ∩ ∩ Hk) ∩ Hk+1 is a finite index by part (a) of the preceding exercise.
35. a. a ≡ a for all a because a = 1a1. If a ≡ b, then a = hbk, h H, k K, so b = h−1ak−1, that is b ≡ a. If a ≡ b and b ≡ c, then a = hbk, b = h1ck1, so a = (hh1)c(kk1). Thus a ≡ c.
H, k K, so b = h−1ak−1, that is b ≡ a. If a ≡ b and b ≡ c, then a = hbk, b = h1ck1, so a = (hh1)c(kk1). Thus a ≡ c.
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