### More conjugation shenanigans

After I wrote my last post on the condition in a group, I had a sudden doubt concerning the case in which this arose: there we assume that we have a coset such that for all . I claimed that this implies , but really the argument I wrote just means that : for all , we get , hence . But what about the other inclusion?

I had in mind a case where the groups involved are finite, or close enough, so that the reverse inclusion is indeed obviously true. But it is amusing to see that in fact what I wrote is correct in all cases: if is a subgroup of an arbitrary group and satisfies , then in fact : taking the inverse of the inclusion gives

I find this interesting because, when it comes to the normalizer, the analogue of this fact is not true: the condition is not, in general, equivalent with .

(I found this explained in an exercise in Bourbaki, Algèbre, Chapitre I, p. 134, Exercice 27, or page 146 of the English edition; here is a simple case of the counterexample from that exercise: consider the group of permutations of ; consider the subgroup which is the pointwise stabilizer of , and the element which is just

Then we have , because the left-hand side is the pointwise stabilizer of which is a subset of . But is not equal to , because is the pointwise stabilizer of , and there are elements in which fix but not .)

It seems natural to ask: what exactly is the set of all such that is equivalent to ? This set contains the line for , and also the coordinates axes and (since we then deal with cosets of ).

In fact, one can determine : it is the complement of the line , for (which corresponds to conjugation). Indeed, suppose for instance that . Let and be such that .

We can write

and

Since , *and since* is the common exponent of and at the two extremities of the right-hand side, it follows that this right-hand side is contained in .

A similar argument works for , using

where again it is crucial that the coefficient appears on both sides, and that it is .

Since we already know that , and in fact that for (the same setting as the counterexample above works, because the we used is an -th power for every ), we have therefore determined the set …

Johan — 04.05.2014 @ 2:25

Fun!