Question

This is just a question originated from some random thoughts. I hope it's nevertheless fit for mo.

It's possible to find a proper subgroup of $GL(n,\mathbb{C})$ isomorphic to $GL(n,\mathbb{C})$ itself (simply as set-theoretical groups, not algebraic groups): just embed $\sigma:\mathbb{C}\hookrightarrow\mathbb{C}$ by a map which is identity on algebraic numbers and is "a shift" on a trascendence basis; then take invertible matrices with entries in $\sigma(\mathbb{C})$.

The ring $\mathbb{Z}$, instead, doesn't admit an injective non surjective morphism into itself, so the above trick does not apply to the following question:

Does $GL(n,\mathbb{Z})$ have any proper subgroup which is isomorphic to $GL(n,\mathbb{Z})$ itself?

My guess is "no", but I'm not sure the following reasoning makes sense (I must say I have no experience with group schemes over a non-(algebraically colsed field)). Suppose you have a proper subgroup $G$ with the above property. Thinking of $GL(n,\mathbb{Z})$ as a group scheme, it should be possible (...is it?...) to put on $G$ the structure of a closed subgroup-scheme, and it will still be (...will it?...) isomorphic to $GL(n,\mathbb{Z})$ now as a group scheme. Then base change this to a field, say $\mathbb{C}$, obtaining a closed algebraic subgroup of $GL(n,\mathbb{C})$ isomorphic to the ambient, which is impossible by dimension.

(btw, I've the feeling that the question will be readily answered by a much simpler -and correct- argument by some user!)

Answer 1

For $n=2$, we have $PSL(2,\mathbb Z) = \mathbb Z_2 * \mathbb Z_3 = \langle u , v | u^2 = v^3 = 1 \rangle$. The map $u \mapsto (uv)^n u (uv)^{-n}, v \mapsto (uv^2)^m v (uv^2)^{-m}$ for $m, n$ large enough provides a non surjective morphism of $PSL(2, \mathbb Z)$ onto itself.

Answer 2

If $n>2$, this is a particular case of the main result in [G. Prasad, Discrete subgroups isomorphic to lattices in semisimple Lie groups, Amer. J. Math. 98 (1976), no. 1, 241--261], namely irreducible lattices in linear semisimple Lie groups are co-Hopf (where a group is called co-Hopf if it is not isomorphic to its proper subgroup). I think, $GL(2,\mathbb Z)$ is also co-Hopf, but at the moment I am not sure how to prove this; note that $GL(2,\mathbb Z)$ has $\mathbb Z*\mathbb Z$ as a finite index subgroup, and $\mathbb Z*\mathbb Z$ isn't co-Hopf.

EDIT: As I mentioned in comments, Prasad's paper implies the result when the ambient Lie group is semisimple, which covers the cases of $SL(n,\mathbb Z)$ and $PGL(n,\mathbb Z)=PSL(n,\mathbb Z)$ when $n>2$. I cannot find a cheap proof for $GL(n,\mathbb Z)$, but here is an ad hoc argument.

A key point is that any injective endomorphism $\phi$ of $GL(n,\mathbb Z)$ must have finite cokernel (i.e. its image has finite index). Indeed, its restriction to $SL(n,\mathbb Z)$ followed by projection $GL(n,\mathbb Z)\to PGL(n,\mathbb Z)$ is a homomorphism of lattices $\phi_0: SL(n,\mathbb Z)\to PSL(n,\mathbb Z)$ in locally isomorphic semisimple Lie groups so Margulis superrigidity implies that $\phi_0$ has finite cokernel, and hence so does $\phi$. Now if $-I_n$ is not in the image of $\phi$, then $GL(n,\mathbb Z)$ embedds as a finite index subgroup into $PGL(n,\mathbb Z)$, so by $GL(n,\mathbb Z)$ is isomorphic to a lattice in a semisimple Lie group, hence it is co-Hopf by Prasad. If $-I_n$ lies in the image of $\phi$, then $-I_n=\phi(-I_n)$, so $\phi$ descends to an injective endomorphism of $PGL(n,\mathbb Z)$, which by Prasad is onto, and it easily implies that $\phi$ is onto.

Answer 3

For $G=GL(2,\mathbb Z)$ there is no proper subgroup isomorphic to it. Consider the dihedral group $D$ of isometries of a regular 6-gon. There is only one conjugacy class in $G$ of subgroups isomorphic to $D$. Indeed given such a subgroup, simply call it $D$, one may adapt the inner product to it, by averaging, so as to make $D$ consist of orthogonal matrices. Then take a basis of $\mathbb Z^2$ consisting of shortest vectors making an obtuse angle, say. Let $s$ be the element that swaps the two basis vectors. We now look for an element $u$ of order four with $susu=1$ so that $u^2$ commutes with the elements of $D$. There is very little choice and we find that $u$ together with $D$ generates $G$.