|Aut(G)| dependent on orders of elements of G?

Question

I'm trying to prove a theorem, but I feel completely stuck. Is it true that if all elements of $G$, which is abelian, have order less than or equal to $2$, then $|Aut(G)|$ must be odd? And if so, then could you please provide with some insight? Once I get this I will probably be able to move further with my theorem.

Answer 1

Since every non-trivial element of $G$ is of order 2, then by the Cauchy's theorem $G$ is of order $2^n$ for some $n$. Then you should prove that $Aut(G)$ is isomorphic to $GL(n,2)$, the group of all binary $n\times n$ matrices. The order of this group equals $(2^n -1)(2^n-2)(2^n-4)...(2^n-2^{n-1})$. For example if $n=2$ then it would be just Alan's example and $|Aut(G)| = 6$. I think you can easily see that this number is odd if and only if $G \cong \mathbb{Z}_2$.

Answer 2

This is not true. The Klein 4 group has all elements of order 2 or 1, but the automorphism group of the Klein 4 group is $S_3$ which has order 6