Epimorphism from GL(2,Z) to GL(2,Z)

Question

Is there an epimorphism $f\colon \mathrm{GL}(2,\mathbb{Z})\to \mathrm{GL}(2,\mathbb{Z})$ which is not injective? Here, $\mathrm{GL}(2,\mathbb{Z})$ is the group of invertible $2\times 2$ matrices with integer entries.

Added. The answer is no: this group is Hopfian. I know a proof in which it is shown that this group is finitely generated (easy) and residually finite(tricky). I suspect that there exists another more "elementary" proof if we use isomorphisms of this group.

I'd really like to see such a proof and so I would be thankfull if someone wrote it (if there exists of course).

Answer 1

No. It is a theorem of Maltsev (or Malcev depending on how the name is translated into Latin characters) that every finitely generated subgroup of $GL(n,\mathbb{C})$, i.e. a linear group, is residually finite and, hence, Hopfian, which means that every epimorphism is injective.

Answer 2

The Maltsev proof is in fact quite easy. Let me give the proofs of the two relevant results.

Theorem: If $G$ is finitely generated and residually finite then $G$ is Hopfian, ie every epimorphism is an isomorphism.

Proof: Suppose not; let $f:G\to G$ be an epimorphism with some non-trivial $g$ in its kernel. For each $n$ let $g_n$ be such that $f^n(g_n)=g$. Let $q: G\to Q$ be a map to a finite group such that $q(g)\neq 1$. Now $q\circ f^n$ kills $g_{n+1}$ but not $g_n$, so the homomorphisms $q\circ f^n$ are all distinct. But there are only finitely many homomorphisms from a finitely generated group to a finite group, so this is a contradiction. QED

A Mariano indicates, the proof that $GL_2(\mathbb{Z})$ is residually finite is also easy.

Theorem: $GL_2(\mathbb{Z})$ is residually finite.

Proof: Let $A\in GL_2(\mathbb{Z})$, thought of as a matrix. Choose some prime $p$ that does not divide all of the entries of $A-I$. The reduction homomorphism $GL_2(\mathbb{Z})\to GL_2(\mathbb{Z}/p)$ is a map to a finite group that does not kill $A$. QED

See - easy!