Characterization of elements of prime order in a symmetric group

Let p be a prime. Show that an element has order p in S_n if and only if its cycle decomposition is a product of commuting p-cycles. Show by an explicit example that this need not be the case of p is not prime.

(\Leftarrow) If \sigma \in S_n is a product of commuting p-cycles, then |\sigma| = p by the lemma to the previous exercise.

(\Rightarrow) Let \sigma \in S_n be an element of order p, and suppose that \sigma is not aa product of commuting p-cycles. Then the cycle decomposition of \sigma contains an n-cycle \tau for some n \neq p, and we can write \sigma = \tau \omega for some \omega \in S_n. Note that if n > p, then |\tau| = n > p by a previous exercise. Then, also by a previous exercise, \sigma^p = \tau^p \omega^p \neq 1, a contradiction. Now if 1 < n < p, then \mathsf{gcd}(n,p) = 1. So by a previous theorem, \tau^p is an n-cycle, thus not 1. So \sigma^p = \tau^p \omega^p \neq 1, also a contradiction. So no such \tau exists in the decomposition of \sigma, so that \sigma is a product of commuting p-cycles.

Consider \sigma = (1\ 2\ 3)(4\ 5). Now \sigma is an element of order 6 in S_n for n \geq 5, but has cycle shape (3,2). Thus the result does not hold (for n \geq 5) if p is not prime.

The result holds vacuously in S_1, S_2, and S_3, since the possible cycle shapes in these groups are (1), (2), and (3). (There are no elements of composite order.)

In S_4, the result holds for composite p in an uninteresting way: the cycle shapes in S_4 are (1), (2), (2,2), (3), and (4). The only elements of composite order have cycle shape (4), and are thus length 1 products of commuting 4-cycles.

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  • Bobby Brown  On August 30, 2010 at 8:04 pm

    Unless I am overlooking something, you left out possibly the most challenging part of the problem: the explicit example!!!! Maybe it’s not so hard, but I’m not seeing it…thanks!


    • nbloomf  On August 30, 2010 at 8:17 pm

      Thanks for the heads up! It should be fixed now.

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