Result related to Brauer-Fowler theorem?

[sans_galois]

Let G be a finite group with involution x such that the centralizer of x coincides with the subgroup generated by x. Then |G:G'|=2.

I think this has something to do with Brauer-Fowler, but I don't see how the derived subgroup comes up

Comments

[stoutfellow]

[x,a] = [x,b] iff x commutes with (a^-1)b. Hence the map a->[x,a] is 2-1. That implies |G:G'| is at most 2; I don't immediately see why you get equality, though.

[sans_galois]

It has something to do with the degree one representations of G, since |G:G'| is simply a count of such reps. But I can't for the life of me see how CG(x)=< x > relates to the number of degree 1 reps.

[mslawllp]

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