(Group Theory) Is g^[G:A] necessarily in A if A isn't necessarily normal?

Hello,

Fair warning: this is not itself a homework problem but it IS
something which arose while working on a homework problem.

Namely, if G a (maybe infinite) group, A a (maybe nonnormal)
subgroup of G w/finite index [G:A], is it true that for all g in G,
g^[G:A] is in A?

If A is normal, this is true and very trivial even. But if A
isn't necessarily normal...... well, to ME it's hard, at least :)

Some things I've tried... g^i is in A for *some* 0<i<=[G:A].
Proof: Assume not. Look at gA, g^2 A, ..., g^[G:A] A. These are [G:A]
cosets, each can have one of [G:A]-1 values (since the coset 1A is by
assumption not in the list), by pigeonhole g^m A = g^n A with
0<m<n<=[G:A], so g^(n-m) in A contrary to assumption.

We can define a multiplication on the elements g^i A, i integer,
by (g^i A)(g^j A)=g^(i+j) A, and it is seen to be well-defined and to
give rise to group structure (say A') even though A may not be normal.
We can define a map pi: <g> -> A' by pi(g^i)=g^i A and by isomorphism
theorems we get <g>/(Ker pi) ~ A', and of course Ker pi is <g>
intersect A so this could also be written <g>/(<g> intersect A) ~ A'...
but this rambling doesnt seem to lead anywhere..

Thanks in advance for any insight you can shed here.
Snis

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