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Let $G$ be a finite abelian group and $\sigma :G\rightarrow G$ an automorphism of order two ($\sigma\circ \sigma =id_G$). Denote by $F$ and $A$ the subgroups 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 the "square" $\phi: G\rightarrow G$, $\phi(g)=g^2$ is an automorphism (namely every element has a square root), 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?

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    $\begingroup$ In the special case when $G$ is elementary abelian,the only indecomposable modules for the action are the trivial module and the regular module of dimension $2$, so you can describe the action precisely. $\endgroup$
    – Derek Holt
    Commented May 16, 2023 at 10:15
  • $\begingroup$ If $g_1,g_2 \in A$, then $\sigma(g_1)=g_1^{-1}$, $\sigma(g_2)=g_2^{-1}$, and hence $\sigma(g_1g_2)=\sigma(g_1)\sigma(g_2)=g_1^{-1}g_2^{-1}=(g_1g_2)^{-1}$. Also if $g\in A$ $\sigma(g^{-1})=g=(g^{-1})^{-1}$, so $A\subset G$ is a subgroup. $\endgroup$
    – Andrea Antinucci
    Commented May 16, 2023 at 17:47
  • $\begingroup$ Abelian groups with involution (I will write $\sigma$-groups for short) form a variety, hence any $\sigma$-group is a subdirect product of subdirectly irreducible (sdi) $\sigma$-groups. Now, what are these? Recall that sdi abelian groups are subgroups of the Prüfer groups $Z(p^\infty)$ (the finite ones being cyclic groups of prime power order). Every $\sigma$-group $(G,\sigma)$ embeds into $(G\times G,s)$ where $s(x,y)=(y,x)$ via $x\mapsto(x,\sigma(x))$, thus using a subdirect decomposition of $G$, $(G,\sigma)$ embeds into a product of $(Z(p^\infty)\times Z(p^\infty),s)$. ... $\endgroup$
    – Emil Jeřábek
    Commented May 19, 2023 at 9:20
  • $\begingroup$ ... Moreover, it is easy to see that $(Z(p^\infty)\times Z(p^\infty),s)\simeq(Z(p^\infty),x)\times(Z(p^\infty),-x)$ for odd $p$ (I’m using additive notation). Since quotients of $\sigma$-groups uniquely correspond to $\sigma$-subgroups, sdi $\sigma$-groups are closed under $\sigma$-subgroups, and it is straightforward to check that $(Z(p^\infty),\pm x)$ and $(Z(2^\infty)\times Z(2^\infty),s)$ are sdi. Thus, sdi $\sigma$-groups are exactly the $\sigma$-subgroups of $(Z(p^\infty),\pm x)$ and $(Z(2^\infty)\times Z(2^\infty),s)$; the finite ones are $(C_{p^k},\pm x)$ and ... $\endgroup$
    – Emil Jeřábek
    Commented May 19, 2023 at 9:26
  • $\begingroup$ ... $\sigma$-subgroups of $(C_{2^k}\times C_{2^k},s)$. The latter can be classified in more detail, but this is not very illuminating. But in particular, the cyclic ones are $(C_{2^k},\alpha x)$ where $\alpha^2\equiv1\pmod{2^k}$, i.e., $\alpha=\pm1$ or $\alpha=2^{k-1}\pm1$. Any finite $\sigma$-group is thus a finite subdirect product of such finite sdi $\sigma$-groups. I do not know how much a decomposition into a direct product of directly indecomposable $\sigma$-groups will differ, but I suspect there do exist directly indecomposable finite $\sigma$-groups that are not sdi. $\endgroup$
    – Emil Jeřábek
    Commented May 19, 2023 at 9:31

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