Show that for all . Determine the elements in this centralizer explicitly.
First we prove a lemma.
Lemma: If has order 2, then . Proof: If , then . Then . Thus . If , then . If , then , a contradiction. So , hence .
We know that the number of conjugates of in for is the number of products of disjoint 2-cycles. In this previous exercise we found that this number is .
Thus ; using Lagrange’s Theorem and since , we have . By the Lemma, .
Now all the permutations which fix centralize , and there are of these. In addition, and centralize . Thus all elements of the form where fixes centralize and (not both 0); there are such elements. Moreover, note that and , so that centralizes . Thus every element of the form , where fixes , centralizes ; there are such elements. Similarly, all elements of the form centralize , for another , as do and . These elements are all distinct, and total . Thus we have exhausted the centralizer of .