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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?

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3 Answers 3

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.

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Wait....which is your definition of "semisimple"? According to the usual definition $GL(n,\mathbb{C})$ and $GL(n,\mathbb{R})$ are not semisimple, is it right? –  Qfwfq Sep 15 '10 at 13:09
    
(btw, is the subgroup isomorphic to $\mathbb{Z}*\mathbb{Z}$ the one generated by the matrices $(1,1;0,1)$ and $(1,0;1,1)$?) –  Qfwfq Sep 15 '10 at 13:18
    
That $GL_n$ is not semisimple is no problem. Just pass to commutator subgroups to get to the semisimple case. –  Wilberd van der Kallen Sep 15 '10 at 13:37
    
I was hasty. Prasad's paper does imply that $SL(n,\mathbb Z)$ is co-Hopf for $n>2$, but I need to look at the proofs there to be sure if they work for $GL(n,\mathbb Z)$, which factors onto $\mathbb Z_2$ with kerner $SL(n,\mathbb Z)$. I am pretty sure that for higher rank lattices co-Hopf property is inherited by finite extensions but one needs to find a reference. –  Igor Belegradek Sep 15 '10 at 13:39
    
Wouldn't this follow from $SL_n$ case? Suppose $\Gamma$ is a proper subgroup of $GL_n(Z)$ isomorphic to $GL_n(Z)$. Any isomorphism also establishes an isomorphism between the commutator subgroups, which is $SL_n(Z)$ in one case, and a proper subgroup that has to be isomorphic, which is impossible by Igor's comment. I think it should also be possible to deduce this from Superrigidity theorem of Margulis (or a weaker forms of it). Although there are some passings to finite index subgroups in the theorem that may make it hard to apply as coHopfian is sensitive to passing to f.i. index subgroup. –  Keivan Karai Sep 15 '10 at 17:41

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.

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Dear Pierre, how do you explicit the description of $PSL(2,\mathbb{Z})$ as that free product? Maybe, thinking of it as the modular group, is it generated by the moebius transformations $z\mapsto 1/z$ and $z\mapsto -1-1/z$? Thanks –  Qfwfq Sep 15 '10 at 16:17
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Exactly, and looking at the action on the upper half-plane, one checks that there is no relation beside $(z\mapsto 1/z)^2=1$ and $(z\mapsto -1-1/z)^3=1$. –  Pierre Dehornoy Sep 15 '10 at 17:44
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The trick of Pierre easily extends to $SL(2,\mathbb Z)$. Recall that in that case the relations are $u^4=v^6=1$ and $u^2=v^3$. See picture on page 35 of Serre's book on trees. For $m$, $n$ large enough, the map $u\mapsto (uv)^nu(uv)^{-n}, v\mapsto (u^2v)^mv(u^2v)^{-m}$ provides a non surjective injective endomorphism of $SL(2,\mathbb Z)$ . –  Wilberd van der Kallen Sep 16 '10 at 8:04

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 $G$.

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