If $G$ is finite abelian and there exists an element of order $> 2$ then $|Aut(G)|$ is even

Question

I would appreciate if you could please evaluate my proof and point out any mistakes I made.

Proof:

Since $G$ is abelian, $Inn(G) \cong \{1\}$, thus $Aut(G) = Out(G)$. But the only automorphisms possible are permutations of the elements of $G$. In addition, every automorphism must preserve the identity. Suppose $|G|=n > 2$. Therefore, $|Aut(G)|=\frac{(n-1)!}{2!}$, where $(n-1)!$ is even since all factorials $>1$ are even.

Answer 1

Unfortunately, your attempt to a proof is completely wrong. You can't just count permutations of $G$ that fix $1$, because they aren't generally automorphisms. Only some permutations fixing $1$ are.

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The map $\iota\colon G\to G$ defined by $\iota(x)=x^{-1}$ is an automorphism of $G$ if and only if $G$ is abelian.

Clearly $\iota\circ\iota=\mathit{id}_G$ is the identity.

Let $x\in G$ have order $n>2$. Then $\iota(x)=x^{n-1}\ne x$ and so $\iota\ne\mathit{id}_G$.

Therefore $\iota$ has order $2$. So $\operatorname{Aut}(G)$ has a subgroup of order $2$.

Note that the finiteness of $G$ is irrelevant.

Answer 2

let $i(g)=g^{-1}, F:Aut(G)\rightarrow Aut(G)$ defined by $F(f)=i\circ f$, $F^2$ is the identity. Suppose $F(f)=f$, let $x\in G, x^n=1, n>2$ and $f(u)=x, F(f)(u)=i(f(u))=i(x)=x^{n-1}\neq x$ contradiction. So $F$ is an involution of $Aut(G)$ without fixed point, thus the cardinal of $Aut(G)$ is even.