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How can we calculate effectively the subgroups of $G: = \mathrm{GL}(n, \mathbb Z)$ which fix pointwise a given submodule $S$ of $\mathbb Z^n$ in the action of $G$ on $\mathbb Z^n$ by left multiplication?

Note: Suppose $S$ is freely generated by $\xi_1, \xi_2, \cdots, \xi_s$ then the clearly the problem is equivalent to describing the unimodular matrices fixing each of the tuples $\xi_1, \cdots \xi_s$. If the $\xi_j$s have a simple form, e.g., $\xi_j$s are the standard basis vectors $e_1, \cdots, e_n$ of $\mathbb Z^n$ then the form of matrices fixing each $\xi_j \ (j = 1, \cdots, s)$ is easy to determine. But in general the method to obtain the stabilizer is not clear.

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    $\begingroup$ One can boil down to the case of a set of basis elements as follows: let $S'$ be the inverse image of the torsion subgroup of $\mathbf{Z}^n/S$ in $\mathbf{Z}^n$. Then the pointwise stabilizer of $S$ also fixes pointwise $S'$. And $S'$ is a direct factor of $\mathbf{Z}^n$. Hence by a change of basis (i.e., conjugation by some matrix in $\mathbf{GL}_n(\mathbf{Z})$ we can boil down to $S'$ being generated by the first $k$ vectors, which gives an obvious block-triangular decomposition. $\endgroup$
    – YCor
    Commented Jun 12, 2020 at 10:24
  • $\begingroup$ So the effective part of the question boils down to compute $S'$ and change the basis to get $S'$ as generated by the first basis vectors. $\endgroup$
    – YCor
    Commented Jun 12, 2020 at 10:25
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    $\begingroup$ Algorithmically, you would put the matrix defining $S$ into Smith Normal Form. $\endgroup$
    – Derek Holt
    Commented Jun 12, 2020 at 12:08

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