Saw an exam question quite a while ago on determining whether the statement:
Let $G$ be an abelian group of order $2^n$. Suppose that $G$ has $m$ involutions. Then the isomorphism class of $G$ is uniquely determined.
is true. I was never able to solve it - I said that an element of $G$ must have order $2^i$ for some $i \le n$ by Lagrange's and that if $g$ is of order $2^i$ then $g^{2^{i - 1}}$ is an involution, and so there are only $m$ possible values for $g^{2^{i - 1}}$. I blagged that this meant that the orders of the elements are fixed by knowing the number of involutions - but this does not seem to use the abelian property anywhere.
There doesn't seem to be any counterexamples for $n = 2, 3, 4$. Is the statement true? If not, is this argument along the right lines?
I think the fundamental theorem of finitely generated abelian groups should be useful here. We know that $G$ is a product of cyclic groups with even order. Then it comes down to does the number of involutions in $G$ determine invariant factors of $G$