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This is problem of Rotman's Exercise 7.9(i).

If $G$ is finite abelian group with $|G| >2$, then $\operatorname{Aut}(G)$ has even order.

How can I approach to this problem? Could you suggest some hints?

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up vote 7 down vote accepted

Hint: By Lagrange's theorem, it suffices to find an element of $\operatorname{Aut}(G)$ with even order, or order $2$. Since $G$ is abelian, what do you know about inversion?

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What you mean is that I should show that there is some element $k \in Aut(G)$ such that $|k|=2$. Yes, I know it by lagrange theorem. However, how can I show element of order 2 is? – user124697 Feb 24 '14 at 4:22
Think about the inversion map $g \mapsto g^{-1}$. – user61527 Feb 24 '14 at 4:23
Hmm but I think this map doesn't ensure existence of nonidentity automorphism which has order 2. – user124697 Feb 24 '14 at 4:25
@user12469 Why not? There are three things to verify: It needs to be an automorphism (do you see why abelian is important here?), it needs to not be the identity (do you see why the group order not being $1$ or $2$ is important here?), and it needs to have order $2$. – user61527 Feb 24 '14 at 4:27
In the extreme case that $G$ is an $\mathbb F_2$-vector-space, taking (additive) inverse is identity; in this case you need a special argument. – Lubin Feb 24 '14 at 4:28

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