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

If is a product of commuting -cycles, then by the lemma to the previous exercise.

Let be an element of order , and suppose that is not aa product of commuting -cycles. Then the cycle decomposition of contains an -cycle for some , and we can write for some . Note that if , then by a previous exercise. Then, also by a previous exercise, , a contradiction. Now if , then . So by a previous theorem, is an -cycle, thus not 1. So , also a contradiction. So no such exists in the decomposition of , so that is a product of commuting -cycles.

Consider . Now is an element of order 6 in for , but has cycle shape . Thus the result does not hold (for ) if is not prime.

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

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

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## Comments

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!

Bob

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