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First, let $A$ be a finitely generated free abelian group, and $s$ an automorphism of order $2$ of $A$. Set $G=\{1,s\}$. Then we know that $A$ is a sum of indecomposable $G$-lattices $A_i$, where each $A_i$ is either $(\mathbb{Z},1)$, or $(\mathbb{Z},-1)$, or $(\mathbb{Z}+\mathbb{Z},\sigma)$, where $\sigma$ interchanges the basis elements of $\mathbb{Z}+\mathbb{Z}$. Thus we know a "canonical form" of $(A,s)$.

Now let $A$ be a finite abelian $p$-group, where $p$ is a prime number (for example, $p=2$). Let $s$ be an automorphism of order $2$ of $A$. As before, set $G=\{1,s\}$. What is a "canonical form" of $(A,s)$? What are the indecomposable $G$-modules here?

If the question is too difficult, then at least what are interesting examples of indecomposable $G$-modules?

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There is a big difference between the case $p=2$ and $p\neq 2$. If $p\neq 2$ then $A$ will be the direct sum of $(\mathbb{Z}/p^n,1)$ and $\mathbb{Z}/p^n,-1)$ for some $n$. In this case all the indecomposables are also irreducible.

If $p=2$ and $A$ is 2-elementary abelian, then $A$ will split as the direct sum of $(\mathbb{Z}/2,1)$ and $(\mathbb{Z}\oplus\mathbb{Z},\sigma)$ where $\sigma$ interchanges the basis elements. In this case the first indecomposable is also irreducible, and the second one is not.

In case $A$ is a non elementary abelian 2-group, the situation is more complicated, and we can have many indecomposable modules. Here is how to construct such modules: Let $B=(b_{i,j})$ be an $n\times n$ matrix over $\mathbb{Z}/2$. Let $A = \mathbb{Z}/4^n$ with basis $\{e_i\}$ and let the action of $s$ be given by $s(e_i) = e_i + 2\sum_j b_{i,j}e_j$ Then the module $A$ will be indecomposable if and only if the matrix $B$ is indecomposable (that is- it is not conjugate to any nontrivial block diagonal matrix). Moreover, the conjugacy class of $B$ is an invariant of the isomorphism class of $A$. So non conjugate indecomposable matrices will give non isomorphic indecomposable modules. Since there are many such conjugacy classes, we get a variety of indecomposable modules.

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    $\begingroup$ I believe Mikhail Borovoi also wants to allow finite length $\mathbb{Z}_p$-modules, not only finite length $\mathbb{Z}/p\mathbb{Z}$-modules. $\endgroup$
    – Jason Starr
    Commented Aug 3, 2015 at 12:20
  • $\begingroup$ The situation there is slightly more complicated, because one should consider all the possible quotients $\mathbb{Z}/p^n$. In case $p\neq 2$ All the indecomposables are $(\mathbb{Z}/p^n,1)$ and $(\mathbb{Z}/p^n,-1)$. I believe that the case $p=2$ will be slightly more complicated. $\endgroup$
    – Ehud Meir
    Commented Aug 3, 2015 at 12:48
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    $\begingroup$ Actually, in case $p=2$ you can have very special indecomposable modules. Let me give an example: Take a $n\times n$ matrix $B=(b_{i,j})$ over $\mathbb{Z}/2$. Take now $A = (\mathbb{Z}/4)^n$ with basis $e_i$, and define the action of $s$ to be $s(e_i) = e_i + \sum_j 2b_{i,j}e_j$ (where $2\cdot 0 = 0$ and $2\cdot 1 = 2 mod 4$). Then one can show that this module is indecomposable if and only if the matrix $B$ is indecomposable. Since for every $n$ we will have an indecomposable matrix of size $n\times n$, we will get a very rich family of indecomposable modules. $\endgroup$
    – Ehud Meir
    Commented Aug 3, 2015 at 13:38
  • $\begingroup$ Just one more remark: By an indecomposable matrix I mean a matrix which cannot be written nontrivially as a block diagonal matrix (after conjugation) $\endgroup$
    – Ehud Meir
    Commented Aug 3, 2015 at 13:44
  • $\begingroup$ Thank you, Ehud. I would appreciate if you include this in your answer. $\endgroup$
    – Mikhail Borovoi
    Commented Aug 3, 2015 at 14:12

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