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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    $\begingroup$ @markvs I think your hint is far too sloppy to be taken seriously in consideration. Have you thought of the case when the groups have order 2? $\endgroup$
    – YCor
    Mar 13, 2022 at 14:37
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    $\begingroup$ @YCor thanks for the suggestion. I hadn't thought of that case. I think my statement is also true in this case due to a theorem of Hua and Reiner: "Every matrix $A\in M_n(\mathbb{Z})$ s.t. $A^2=I_n$ is integrally similar to a matrix of the form $W(x,y,z)=\underbrace{L\oplus\cdots\oplus L}_{x} \oplus (-I)_y \oplus I_z$, where $2x+y+z=n$ and $L=\begin{pmatrix} 1&0\\1&-1 \end{pmatrix}$". Therefore, $H_1$ and $H_2$ being non-conjugated implies that $h_1 \sim W$ and $h_2\sim W'$ with $W\neq W'$. Then $I_1 \oplus W \not\sim I_1 \oplus W'$ $\endgroup$ Mar 13, 2022 at 18:50
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    $\begingroup$ @AlejandroTolcachier thanks. Indeed this settles the case provided $(x,y,z)$ is unique. But it's indeed unique: $2x+y+z$ is fixed (equal to $n$), $z-y$ is determined as the trace, and $x$ is the number of Jordan blocks of size $\ge 2$ in the reduction mod $2$. $\endgroup$
    – YCor
    Mar 13, 2022 at 18:59
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    $\begingroup$ If $H$ and $H^{\prime}$ each fix no non-zero element of $\mathbb{Z}^{n}$, then the fact that $hZ = Z$ for all $h \in H$ forces $Z = [0]$, and then $W$ is indeed invertible, so what you want does follow ( $Z$ and $W$ as in the body of the question). $\endgroup$ Mar 16, 2022 at 15:55
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    $\begingroup$ Over $\mathbb{Z}_p$ I think the question has an affirmative answer, since the category of $\mathbb{Z}_p$ representations of $H_1\cong H_2$ is Krull-Schmidt, so adding copies of the trivial rep doesn’t change isomorphism class. One could maybe do something local/global from this.. $\endgroup$
    – Chris H
    Mar 16, 2022 at 22:20

1 Answer 1


$\def\ZZ{\mathbb{Z}}\def\GL{\text{GL}}$We can make partial progress using:

Warfield, R. B. jun., Cancellation of modules and groups and stable range of endomorphism rings, Pac. J. Math. 91, 457-485 (1980). ZBL0484.16017.

In particular, we can show that if $H_1$ and $H_2$ are conjugate in $GL(n+m)$ then they are conjugate in $GL(n+1)$.

We first rephrase in the language of modules. Let $H$ be the abstract abelian group which is isomorphic to both $H_1$ and $H_2$. Let $R$ be the group ring $\ZZ[H]$. Embedding $H$ into $\GL_n(\ZZ)$ is equivalent to equipping $\ZZ^n$ with the structure of an $R$-module; call our $R$-modules $M_1$ and $M_2$. Conjugating $H_1$ to $H_2$ is equivalent to giving an isomorphism of $R$-modules.

I'll write $A$ for $\ZZ$ considered as an $R$-module with the trivial action of $R$. The hypothesis that $H_1$ and $H_2$ are conjugate in $\GL_{n+m}(\ZZ)$ means that $A^{\oplus m} \oplus M_1 \cong A^{\oplus m} \oplus M_2$. So your question is:

Given $M_1$ and $M_2$ are $R$-modules, both of which are free as rank $n$ as $\ZZ$-modules, and $A \oplus M_1 \cong A \oplus M_2$, do we have $M_1 \cong M_2$?

Warfield tells us that we should consider the endomorphism ring $E:=\text{Hom}_R(A,A)$. In this case, $E = \ZZ$. We should then compute the stable rank of $E$: The stable rank of $\ZZ$ is known to be $2$. (This also follows from Theorem 4.1 in Warfield, taking $R=S$ as here and $A$ as here.)

Warfield then tells us that, if $E$ has stable range $2$ then $A$ obeys $2$-substitution. In Theorem 1.3, Warfield tells us that $2$-substitution has the following consequence (his $X'$ is my $M_1$ and his $Y$ is my $A \oplus M_2$):

If $A^2 \oplus M_1 \cong A^2 \oplus M_2$ then there is some module $L$ such that $A \oplus M_1 \cong L \oplus M_2$ and $A^2 \cong A \oplus L$.

In our special case, I claim that $L$ must be $A$. To see this, use the hypothesis that $A^2 \cong A \oplus L$. Looking at ranks as $\ZZ$-modules, we must have $L \cong \ZZ$ as a $\ZZ$-module. But $L$ is supposed to be a submodule of $A^2$, which is just $\ZZ^2$ with the trivial $H$-action. So the $H$-action on $L$ is likewise trivial and we have $A \cong L$.

So we have the simpler statement

If $A^2 \oplus M_1 \cong A^2 \oplus M_2$ then $A \oplus M_1 \cong A \oplus M_2$.

This means that conjugacy in $\GL_{n+2}(\ZZ)$ implies conjugacy in $\GL_{n+1}(\ZZ)$.


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