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).

  • $\begingroup$ Please try to use mark-up. It makes things easier to read. $\endgroup$ Jan 11, 2011 at 18:16
  • 1
    $\begingroup$ Why is it tricky to show that $\operatorname{GL}(2,\mathbb Z)$ is residually finite? You automagically have lots of surjections $\operatorname{GL}(2,\mathbb Z)\twoheadrightarrow \operatorname{GL}(2,\mathbb Z_p)$ for all primes $p$, for example... $\endgroup$ Jan 11, 2011 at 18:51
  • $\begingroup$ I mean it's the tricky part of this proof. $\endgroup$
    – t.k
    Jan 11, 2011 at 19:03
  • $\begingroup$ I think Mariano's remark means you don't need to know $GL(2,\mathbb{Z})$ is finitely generated. $\endgroup$
    – user641
    Jan 12, 2011 at 10:11
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    $\begingroup$ Regarding other possible proofs: there's another proof that free groups are Hopfian, using Nielsen transformations. It's given in Lyndon & Schupp. Now, $GL_2(\mathbb{Z})$ is virtually free. It may be possible to use these two facts to come up with a different proof, though I don't know how. $\endgroup$
    – HJRW
    Jan 13, 2011 at 4:12

2 Answers 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!


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.

  • $\begingroup$ That's the proof I already know of. $\endgroup$
    – t.k
    Jan 11, 2011 at 18:08

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