Let $G$ be a finite abelian group and $\sigma :G\rightarrow G$ an automorphism of order two ($\sigma\circ \sigma =id_G$). Denote be $F$ and $A$ the subgroup of fixed and anti-fixed points of $\sigma$ respectively: $$ F=\left\{ g\in G| \sigma (g)=g\right\} \, , \ \ A=\left\{ g\in G| \sigma (g)=g^{-1}\right\} $$ If $|G|$ is odd it is simple to prove that $G=F\times A$. Indeed $|G|$ is odd if and only if $\phi: G\rightarrow G$, $\phi(g)=g^2$ is an automorphism, then $x=\phi^{-1}(g\sigma(g))$ is clearly in $F$, and a simple computations shows that $y=x^{-1}g\in A$.

My question is what about the case in which $|G|$ is even? The result above cannot be true since $A$ and $F$ could have non-trivial intersection, contained in the subgroup of order-2 elements (the kernel of $\phi$). But is there some kind of splitting structure we can prove on $G$, maybe by taking some quotient by the subgroup of order-2 elements?