# Automorphisms of products of $GL_n(\mathbb{Z})$ 's

It is a Theorem of Hua y Reiner (1951) that the group or outer automorphisms $Out(GL_n(\mathbb{Z}))$ is either isomorphic to $\mathbb{Z}/2$, if $n$ odd or $n=2$, or to $\mathbb{Z}/2 \times \mathbb{Z}/2$, if $n>2$ even. I was wondering if anybody knows if there is a reference that deals with the case of the outer automorphisms of

$$\Gamma=\prod \limits_{i=1}^{m} GL_{n_i}(\mathbb{Z}).$$

I would like to know if $Out(\Gamma)$ is finite, and if so, what its order is. Thanks a lot.

• Yes it's always finite. The simplest case is when all $n_i$ are odd, because they're then centerless, and hence any automorphism permutes the product decomposition; if $q_j$ is the number of occurrences of $j$ among the $n_i$, it is then an exercise to check that the Out has order $\prod_j 2^{q_j}q_j!$. If we allow even ones, we have at least (restrictiing to automorphism permuting the factors) $\prod_{j\text{ odd or 2}}2^{q_j}q_j!\prod_{j\text{ even }>2}4^{q_j}q_j!$ and I haven't checked if there's more (maybe there's some mess coming from homomorphisms onto the group on 2 elements). – YCor Aug 11 '17 at 16:39
• @YCor, I can't seem to see it: I agree that "they're then centerless" when all $n_i$ are odd, but why "hence any automorphism permutes the product decomposition"? – LSpice Aug 11 '17 at 17:04
• @LSpice it's a general fact about direct decompositions of centerless groups (of course I forgot to invoke that $SL_n(Z)$ is indecomposable) see Proposition 2 here: normalesup.org/~cornulier/DirDec.pdf and the following "consequence" next page (sorry, in French). There are certainly earlier references. – YCor Aug 11 '17 at 17:19
• Oh by the way for odd $n$ I was thinking of $SL_n(Z)$. For odd $n$, $GL_n(Z)$ is just the direct product $SL_n(Z)\times (Z/2Z)$ so of course if we consider a product of $k$ such guys, $GL_k(Z/2Z)$ should pop out. – YCor Aug 11 '17 at 17:21
• @YCor, sorry again, but shouldn't $\mathrm{GL}_n(\mathbb Z)$ be $\mathrm{SL}_n(\mathbb Z) \rtimes \mathbb Z/2\mathbb Z$ (semi-direct, not direct, product) for $n$ odd? Maybe I just don't know how you're embedding $\mathbb Z/2\mathbb Z$. – LSpice Aug 11 '17 at 19:45

## 1 Answer

Here's a proof of finiteness.

First, it's a general fact that for any centerless directly indecomposable groups $A_1,\dots,A_k$, any automorphism permutes the $A_i$. References are welcome; one (in French), probably much too recent, is Proposition 2 p9 here (2006 paper by myself and Pierre de la Harpe). In particular, if all $A_i$ have finite Out, so does the product.

This applies to $\Lambda=\prod_{i=1}^m\mathrm{PSL}_{n_i}(\mathbf{Z})=[\Gamma,\Gamma]/Z(\Gamma)$.

It follows that some finite index subgroup of $\mathrm{Aut}(\Gamma)$ acts by inner automorphisms on $\Lambda$. To conclude, one has to show that the subgroup $F$ of $\mathrm{Aut}(\Gamma)$ acting as the identity on $\Lambda$ is finite. To show this, it is enough to show that the following finite index subgroup of $F$ is itself finite: the kernel $F'$ of the $F$-action on $Z(\Gamma)\times (\Gamma/[\Gamma,\Gamma])$. The $F'$-action on $[\Gamma,\Gamma]$ consists is by automorphisms of the form $g\mapsto gs(g)$ where $s$ are homomorphisms $[\Gamma,\Gamma]\to Z(\Gamma)$. Since $Z(\Gamma)$ is finite and $[\Gamma,\Gamma]$ is finitely generated, the set of such homomorphisms is finite. Hence some finite index subgroup $F''$ of $F'$ acts trivially on $[\Gamma,\Gamma]$. In turn, the $F''$-action on $\Gamma$ is by automorphisms that are the identity on the finite index subgroup $[\Gamma,\Gamma]$, and have the form $x\mapsto xu(x)$ where $u$ is a map from $\Gamma$ to $[\Gamma,\Gamma]$. Since it's an automorphism, we have $u(xy)=y^{-1}u(x)yu(y)$ for all $x,y$, and $u=1$ on $[\Gamma,\Gamma]$. Combining, we see that $u$ factors through $\Gamma/[\Gamma,\Gamma]$. So $u(xy)=u(yx)$ for all $x,y$. Taking $y$ in $[\Gamma,\Gamma]$ then yields $u(x)=y^{-1}u(x)y$ for all $x$; thus $u$ takes values in $Z([\Gamma,\Gamma]$, which in our case is central in $\Gamma$. So $u$ is a homomorphism $\Gamma\to Z(\Gamma)$. Again the set of such homomorphisms is finite, and we have proved finiteness of $F''$, hence of $F$.

• PS if (and only if) $n_i\ge 3$ for all $i$ then a stronger result is true: every finite index subgroup has finite Out. This follows from Mostow rigidity and some little further arguments. – YCor Aug 11 '17 at 18:48