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By Huntington E.V.

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Now write α = α1 α2 . . αs where each αi is a 2-cycle. Then αγα−1 = α1 α2 . . (αs γαs ) . . α2 α1 since each αi has order 2. Now by Exercises 50 we can replace αs γαs by a cycle the same length as γ. Repeated use of this argument finishes the proof. 52. Say β can be written with m 2-cycles and α with n Then β −1 αβ can be written with 2m + n 2-cycles. 2-cycles. 53. Examining Sn for each n up to 8 as in Example 4 we see that Sn has no element of order greater than 2n. For n = 9 we have |(12345)(6789)| = 20.

38. In D6 , take a = R60 and b = R120 . Then |a2 | = |b2 | but |a| = |b|. 39. If T and U are not closed, then there are elements x and y in T and w and z in U such that xy is not in T and wz is not in U . It follows that xy ∈ U and wz ∈ T . Then xywz = (xy)wz ∈ U and xywz = xy(wz) ∈ T , a contradiction since T and U are disjoint. 40. If a has order p, then a has p − 1 elements of order p. So, there is an element b not in a that also has order p. Then a ∪ b has 2(p − 1) > p elements of order p. 41.

If A, B ∈ H, then det (AB −1 ) = (det A)(det B)−1 is rational. H is not a subgroup when det A is an integer, since det A−1 = (det A)−1 is an integer only when det A = ±1. 36. Let H be the nontrivial proper subgroup of G and pick g ∈ G so that g ∈ H. Then G = g . 3 and Exercise 42 of Chapter 4. 37. First suppose that G is not cyclic. Choose x = e and choose y ∈ x . Then, since G has only two proper subgroups G = x ∪ y . But then xy ∈ y so that x ⊆ y and therefore G = y and G is cyclic. To prove that |G| = pq, where p and q are distinct primes or |G| = p3 , where p is prime we first observe that G is not infinite since an infinite cyclic group has infinitely many subgroups.

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