Let be a conjugacy class and suppose that .

- Show that does not commute with any odd permutation if and only if the cycle type of consists of distinct odd integers.
- Deduce that is a union of two -conjugacy classes if and only if the cycle type of consists of distinct odd integers.

[Hint: Assume first that does not commute with any odd permutation. Observe that commutes with each cycle in its own cycle decomposition, so that each cycle must have odd length. If two cycles have the same odd length , find a product of transpositions which interchanges them and commutes with . Conversely, if the cycle type of consists of distinct integers, prove that commutes only with the subgroup generated by the cycles in its cycle decomposition.]

We begin with a remark.

Remark: If is a cycle, then , where is the set of fixed points of and is naturally identified with a subgroup of .

Now to the main result.

Suppose does not commute with any odd permutation. Write , where is some cycle in the decomposition of and commutes with . (Say is the rest of the cycle decomposition of .) Note that , so that commutes with . By our hypothesis, then, is not an odd permutation. Hence has odd length. Since was an arbitrary cycle in the decomposition of , the cycle type of consists of odd integers.

Now suppose some cycles and in the decomposition of have the same length , and write for some which commutes with both and . (Again, say, the rest of the cycle decomposition of .) Now and are conjugate in , and moreover are disjoint. Write and ; that is, define and such that and for all , where indices are taken mod . Let for each , and let ; note that commutes with . Moreover, by a previous exercise, we have by our definition of . Now for each , we have , with indices taken mod . Hence . Moreover, since is a product of disjoint 2-cycles, . Finally, note that is an odd permutation, being an odd product of commuting transpositions. Now , so that commutes with , a contradiction. So no two cycles in the decomposition of have the same length.

Thus the cycle type of consists of distinct odd integers.

Suppose that the cycle type of consists of distinct odd integers. Then fixes at most one element; that is, . By the remark above, . Now consists of only the identity; hence . Because is a product of even permutations, consists only of even permutations. In particular, does not commute with any odd permutation.

Now for the second point.

Let be a conjugacy class and let .

Note that consists of two -conjugacy classes if and only if does not commute with any odd permutation by §4.3 #20, if and only if the cycle type of consists of distinct odd integers by the above argument.

## Comments

In (), how about defining ? Then the conjugation of changes with so that .

I like it.