a. o(g) = 5. Then G = g^k if k = 1, 2, 3, 4.

c. o(g) = 16. Then G = g^k if k = 1, 3, 5, 7, 9, 11, 13, 15.

a. has generators .

2. has generators

a.. We have , so . As to : , , , , , , . Thus .

c. . Here , so . Similarly , , and . Thus is not cyclic.

a. by Theorem 5.

b. by Theorem 5.

9.

10. a. If o(g) = n and o(h) = m, then (gh)^nm = (gⁿ)^m(hⁿ)^m = 1 because gh = hg.

11. a. If G = a where o(a) = n, let g = a^k. Then

13. a. Observe first that g⁻¹ = g if and only if g = 1 or o(g) = 2. Thus all the elements in the product a = g₁g₂ g_n which are not of order 2 (if any) cancel in pairs because G is abelian. Since and since 1 and the elements of order 2 (if any) all square to 1, the result follows.

15. We have a, ab ⊆ a, b by Theorem 10 because a a, b and b a, b . The reverse inclusion follows because a a, ab and b = a⁻¹(ab) a, ab . Similarly, a, b = a⁻¹, b⁻¹ because a⁻¹, b⁻¹ a, b , and a = (a⁻¹)⁻¹ and b = (b⁻¹)⁻¹ are both in a⁻¹, b⁻¹.

16.

17. a. Since X ⊆ Y and Y ⊆ Y , we have X ⊆ Y . But Y is a subgroup, so X ⊆ Y by Theorem 10.

19. We have xy⁻¹ = y⁻¹x and x⁻¹y⁻¹ = y⁻¹x⁻¹ for all x, y X. If

20. If C₆ = a and C₁₅ = b , then (a³, b), (a, b³), (a, b) all have order 30. Since (x, y)³⁰ = (x³⁰, y³⁰) = (1, 1) for all (x, y) in C₆ × C₁₅, these have maximal order.

21. Each element of S₅ factors into cycles in one of the following ways (shown with their orders).

23. a. We have (ghg⁻¹)^k = gh^kg⁻¹ for all k ≥ 1. Hence h^k = 1 if and only if (ghg⁻¹)^k = 1. It follows that o(h) = o(ghg⁻¹) as in Example 10.

24. a. If h is the only element of order 2 in G, then h = g⁻¹hg for all g G since (g⁻¹hg)² = g⁻¹h(gg⁻¹)hg = g⁻¹h²g = g⁻¹g = 1. Thus gh = hg for all g G, that is h Z(G). Note that C₄ = a , o(a) = 4, has such an element: a².

25. Let G = g and H = h where o(g) = m and o(h) = n. Since we have |G × H| = |G| |H| = mn, it suffices to show that o((g, h)) = nm. We have (g, h)^nm = (g^nm, h^nm) = (1, 1). If (g, h)^k = (1, 1), then g^k = 1 and h^k = 1, so m k and n k. But gcd (n, m) = 1, then implies nm k (Theorem 5 §1.2)), so o((g, h)) = mn, as required.

26. a. Write o(gh) = d. Since gh = hg, we have

27. a. If A ⊆ B, then g^a B = g^b , say g^a = g^bq, . Since o(g) =∞, a = qb. Conversely, if a = qb, then g^a B, so A ⊆ B.

29. Write o(g^k) = m. Then (g^k)^n/d = (gⁿ)^k/d = 1^k/d = 1 implies that m (n/d). On the other hand, write d = xk + yn with (by Theorem 3 §1.2). Then (g^k)^m = 1 implies g^dm = (g^km)^x · (gⁿ)^ym = 1, so n dm. If qn = dm, , then , so (n/d) m. This shows (n/d) = m, as required.

31. a. We have a m and b m, so g^m A and g^m B. Thus g^m A ∩ B, whence g^m ⊆ A ∩ B. Conversely, write A∩ B = g^c . Then g^c A, say g^c = (g^a)^x. Since o(g) =∞, this implies c = ax. Similarly, g^c B implies c = by. Thus c is a common multiple of a and b, so m c by the definition of the least common multiple. This implies A∩ B = g^c ⊆ g^m .

32. (1) ⇒ (2). Let H and K be subgroups of G = g where o(g) = pⁿ. By Theorem 9, let H = g^a and K = g^b where a and b are divisors of pⁿ. Since p is a prime, this means a = p^l and b = p^m. If l ≤ m, this says a b, whence K ⊆ H. The other alternative is m ≤ l, so H ⊆ K.

33. If G is cyclic, it is finite (because infinite cyclic groups have infinitely many subgroups). So assume G is not cyclic. Use induction on the number n of distinct subgroups of G. If n = 1, G = {1} is finite. If it holds for n = 1, 2, . . ., k, let H₁ = {1}, H₂, . . ., H_k, H_k+1 = G be all the subgroups of G. If 1 ≤ i ≤ k then H_i ⊆ G so H_i is finite by induction. So it suffices to show G = H₁ ∪ H₂ ∪ ∪ H_k. But if g G, then g ≠ G because we are assuming that G is not cyclic. Hence g = H_i for some i, so g H_i.

35. a. Let and where the p_i are distinct primes and m_i ≥ 0, n_i ≥ 0 for each i. For each i, define x_i and y_i by

37. Let cards numbered 1, 2, 3, . . . be initially in position 1, 2, 3, . . . in the deck. Then after a perfect shuffle, position 1 contains card 1, position 2 contains card n + 1, position 3 contains card 2, position 4 contains cards n+ 2, . . .. In general,

a.