Let be a finite group, an automorphism of . Let be the set of all such that .

(i) If then prove that is abelian.
(ii) If then prove that has an abelian subgroup of index 2.

(i)

First, we claim that if then commutes with . Simply note that and then invert both sides.

Now, fix an element and consider the set . As , it's not hard to see that (in detail, take the complement and use De Morgan's). From the above if is in this set then and commute, hence is a subset of the center of . But the center of is a subgroup of , so its order has to be a divisor of , but since its order is more than it has to be itself. That is, commutes with all of , i.e. is in the center of

Thus , and again since is a subgroup of with an order exceeding , we have and is abelian.

(ii)

First, cannot be abelian, because if it is then for we have and . In that case it's clear would be a subgroup, which is impossible as its order is not a divisor of .

So, let be an element not in the center of but in (such an element certainly exists as ). Similar to (i), consider , which is as before a subset of , and this time. Of course, is not the whole of so it's a subgroup with order exactly , and index . Then .

We claim that is an abelian subgroup. In fact, . Now for any we have , and we can invert both sides as before, hence and commute and we're done.

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Darn group actions on sets are so annoying when the group and the set are the same, or subgroups of another group, or subgroups of subgroups of something else. Sylow qq