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.

  • 1
    $\begingroup$ 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). $\endgroup$ – YCor Aug 11 '17 at 16:39
  • 1
    $\begingroup$ @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"? $\endgroup$ – LSpice Aug 11 '17 at 17:04
  • $\begingroup$ @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. $\endgroup$ – YCor Aug 11 '17 at 17:19
  • $\begingroup$ 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. $\endgroup$ – YCor Aug 11 '17 at 17:21
  • $\begingroup$ @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 $\mathbb Z/2\mathbb Z$. $\endgroup$ – LSpice Aug 11 '17 at 19:45

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 $F$.

  • 1
    $\begingroup$ 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. $\endgroup$ – YCor Aug 11 '17 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.