In my research I came up with the following question:

**Question:** Let $H_1$ and $H_2$ be finite abelian subgroups of $\mathsf{GL}(n,\mathbb{Z})$. Define $$ H_1'=\left\{\begin{pmatrix} I_m &0\\0&h_1\end{pmatrix}\mid h_1\in H_1\right\},\quad\text{and}\quad H_2'=\left\{\begin{pmatrix} I_m &0\\0&h_2\end{pmatrix}\mid h_2\in H_2\right\}.$$
Suppose $H_1'$ is conjugated to $H_2'$ as subgroups of $\mathsf{GL}(m+n,\mathbb{Z})$ (i.e, there exists $\alpha\in \mathsf{GL}(m+n,\mathbb{Z})$ s.t. $H_2'=\alpha H_1' \alpha^{-1}$). Does this imply that $H_1$ is conjugated to $H_2$ as subgroups of $\mathsf{GL}(m,\mathbb{Z})$?

**Some thoughts:** $H_1$ is isomorphic to $H_1'$ and $H_2$ is isomorphic to $H_2'$, so $H_1$ and $H_2$ are at least isomorphic, but of course this does not imply that are conjugated.
On the other hand, writing $\alpha=\begin{pmatrix} X&Y \\ Z&W \end{pmatrix}$, we have that for every $h\in H_1$ there exists $h'\in H_2$ such that
\begin{align*}
\alpha \begin{pmatrix} I_m&0\\0&h\end{pmatrix}=\begin{pmatrix} I_m&0\\0&h' \end{pmatrix} \alpha&\iff \begin{pmatrix} X& Yh \\ Z& Wh\end{pmatrix}=\begin{pmatrix}X&Y\\h'Z&h'W\end{pmatrix}\\&\iff Yh=Y, Z=h'Z,Wh=h'W.
\end{align*} These conditions do not say too much, because $W$ a priori does not have to be invertible.

I've checked carefully (and certainly not in the more efficient way) using GAP and/or MAGMA the case when $n=2$ and for cyclic subgroups of $\mathsf{GL}(3,\mathbb{Z})$ and the answer to my question is affirmative, but I'm afraid there could be some example for much larger dimensions.

**Some update (14/3)**:

$\bullet$ I've checked for subgroups of $\mathsf{GL}(3,\mathbb{Z})$ and $\mathsf{GL}(4,\mathbb{Z})$ (enlarging them with $I_1$ or $I_2$) and the answer is also affirmative.

$\bullet$ In the comments the case when $H_1$ and $H_2$ have order 2 is settled.

I suspect now that it could be true. If it helpful, in this page one can download the finite subgroups of $\mathsf{GL}(n,\mathbb{Z})$ up to $n=6$.

Any idea would be greatly appreciated! Thanks!

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