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Let $F$ be a finitely generated free $\mathbb{Z}$-module on which the group $G$ of two elements acts via group homomorphisms. Let $F'$ be a $G$-invariant submodule. By Smith normal form we know that we can write $F=\oplus_{i=1}^n F_i$ for some free submodules $F_i$ such that $F'=\oplus_{i=1}^n d_iF_i$ for some $d_1|d_2|\cdots|d_n$ where we allow for $d_i=0$ and require the $d_i$ to be pairwise distinct. Is it true that we can choose the $F_i$ to be $G$-invariant subgroups? How unique is such a choice?

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    $\begingroup$ What do you mean by "the group $G$ of two elements"? If you mean the cyclic group of order 2, then it is the same as a $\mathbb Z[C_2]\cong\mathbb Z[t]/(t^2-1)$-module. I guess that there is no difficulty if you invert $2\in\mathbb Z$, but not sure what happens $2$-adically. $\endgroup$
    – Z. M
    Commented Jul 1 at 10:11
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    $\begingroup$ @Z. M.: sorry, what is ambiguous about "the group $G$ of two elements"? $\endgroup$
    – Qiaochu Yuan
    Commented Jul 1 at 17:39
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    $\begingroup$ There is a classification of finitely generated $\mathbb{Z}$-free $\mathbb{Z}G$-modules. There are three indecomposables, namely $\mathbb{Z}$ with the two possible actions, and the regular representation. Every finitely generated $\mathbb{Z}$-free $\mathbb{Z}G$-module is uniquely a sum of these. So maybe a better question is whether there is a canonical form for pairs consisting of such a module and a submodule. I suspect the answer is that there is, but I haven't taken the time to think it through. $\endgroup$
    – Dave Benson
    Commented Jul 1 at 21:24
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    $\begingroup$ I suspect that the intention was the finite simple group of order two. $\endgroup$
    – Lee Mosher
    Commented Jul 1 at 21:46

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In general the $F_i$ cannot be chosen as $G$-invariant subgroups. What about the example of $F=\mathbb{Z}^2$ where the action is given by swapping the coordinates and $F'=\{(a,b)\mid a+b \mbox{ is even}\}$. Then $d_1=1$ and $d_2=2$. But the only $G$-invariant rank-one summands are generated by $(1,\pm 1)$. Their span does not generate $F$.

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    $\begingroup$ Maybe this is really just an issue with the formulation of the Smith normal form. As written it uses direct summands and for $\mathbb{Z}$-modules this is completely fine, but it could also be expressed with an (for $\mathbb{Z}$-modules) equivalent notion (maybe using filtrations, or short exact sequences or something else). To apply it in this slightly changed setting of $\mathbb{Z}[C_2]$-modules, the two notions are not equivalent anymore and the mysterious other one might be better. $\endgroup$
    – HenrikRüping
    Commented Jul 1 at 12:38

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