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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 type of $B$. 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.

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.

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 $(\mathbb{Z}/p,1)$ and $\mathbb{Z}/p,-1)$. In this case all the indecomposables are also irreducible.

If $p=2$ 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.

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 type of $B$. 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.

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 $(\mathbb{Z}/p,1)$ and $\mathbb{Z}/p,=1)$. In this case all the indecomposables are also irreducible.

If $p\neq 2$ 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.

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 $(\mathbb{Z}/p,1)$ and $\mathbb{Z}/p,-1)$. In this case all the indecomposables are also irreducible.

If $p=2$ 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.