TITLE: Proper subgroup of GL(n,Z) isomorphic to GL(n,Z)?
QUESTION [32 upvotes]: 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?

REPLY [9 votes]: 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 $G$.