# Conjugated subgroups in $\mathsf{GL}(m+n,\mathbb{Z})$ implies conjugated subgroups in $\mathsf{GL}(n,\mathbb{Z})$?

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!

• @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?
– YCor
Mar 13, 2022 at 14:37
• @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 $I_1 \oplus W \not\sim I_1 \oplus W'$ Mar 13, 2022 at 18:50
• @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 $2$.
– YCor
Mar 13, 2022 at 18:59
• 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). Mar 16, 2022 at 15:55
• 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.. Mar 16, 2022 at 22:20

$$\def\ZZ{\mathbb{Z}}\def\GL{\text{GL}}$$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 $$\GL_{n+1}(\ZZ)$$.