Re: finite group of odd order has no nontrivial element conjugate to its inverse

Kira Yamato wrote:
> Russell wrote:
> > Kira Yamato wrote:
> >
> >>Need some help in proving that
> >> For G a finite group of odd order, then if there exists g and h in G
> >>such that h^{-1} g h = g^{-1},
> >> then
> >> g = e, the identity element.
> >
> >
> > Someone asked essentially the same thing a couple weeks
> > ago. (IIRC the question then was to prove g^2 = e, but
> > I think you can see why that's the same thing.)
> >
> >>From your eq'n you can easily get
> > g = h^(-1) g^(-1) h
> > so, substitute that value of g back into the original eq'n
> > to get:
> > h^(-2) g(-1) h^2 = g^(-1)
> > and we can substitute the original eq'n into that result to
> > get:
> > h^(-3) g h^3 = g^(-1)
> >
> > Or in general h^(-n) g h^n = g^(-1) for n odd.
> >
> > Now consider the order of h...
> >
>
> So, since G has odd order, h must have odd order too. So, for some odd
> n, we will have h^n = e. Thus,
> g = g^(-1)
> or
> g^2 = e.
>
> So, we found a cyclic subgroup of order 2 {e,g} which contradicts that G
> cannot have subgroups of order which does not divide |G|.
>
> Thanks.

Right. Of course you should really phrase the "in general"
in terms of induction on n, not that this is much of a change.
I do like quasi's hints, too; I'm a little abashed that I
didn't notice the commutation of h^2 with g^(-1) in my
second step, which leads directly to the result by quasi's
nice line of reasoning. Like him, I hope I didn't spoil
the fun for you.

.

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