Consider the group $G=\langle x_1,x_2,x_3|x_1^2,x_2^2,x_3^2\rangle$. Using a slightly modified version of S. Ivanov's proof here that free groups are residually finite, I can show that this group is residually finite. Since it is clearly finitely generated, this implies it is Hopfian.

Proof: Let $x_{i_n}\cdots x_{i_1}$ be a reduced word in $G$, so $i_k\ne i_{k+1}$ for any $1\le k\le n$. Define $f_1,f_2,f_3\in S_{n+1}$ as follows. For $1\le k\le n$, let $f_{i_k}(k)=k+1$ and $f_{i_k}(k+1)=k$, which is well-defined as $i_{k+1}\ne i_k$, and let each $f_i$ fix every other element of $\{1,\ldots,n+1\}$. Then each $f_i^2=\mathrm{id}$, so $f:G\to S_{n+1}$ defined by $f(x_i)=f_i$ is a homomorphism, and $f(x_{i_n}\cdots x_{i_1})(1)=f_{i_n}\circ\cdots\circ f_{i_1}(1)=n+1$ hence is nontrivial. Thus $G$ is residually finite.

However, I find this proof rather unsatisfying, in part because in the context where $G$ arises I don't see a natural interpretation of residual finitude. Is there a more direct way to prove that $G$ is Hopfian, perhaps analogous to direct proofs that finitely generated free groups are Hopfian (see for example prop. 3.5 of Combinatorial Group Theory by Lyndon & Schupp)?

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    $\begingroup$ This group is a free product of finite groups, so it. Is virtually free. Thus, hopfian property for free groups implies hopfian property for the groups you are considering. It also implies residual finiteness. These groups appear frequently in low dimensional topology and geometry since they are fundamental groups of some 2d orbifolds. $\endgroup$
    – Misha
    Dec 25, 2012 at 6:22
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    $\begingroup$ Incidentally, hyperbolic groups are known to be Hopfian, even though they are not known to be residually finite in general. $\endgroup$
    – Ian Agol
    Dec 25, 2012 at 7:36
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    $\begingroup$ @HW indeed I don't see why a group $G$ with a finite index hopfian $H$ should be hopfian. It's true if $G$ has finitely many subgroups of index $n=[G:H]$ (e.g. if $G$ is fg): let $f:G\to G$ be onto, then it induces an self-injection of the set of subgroups of index $n$, hence a permutation. Hence, some power $m$ of $f$ satisfies $(f^m)^{-1}(H)=H$. This easily allows to conclude that $f^m$ is injective, proving $G$ is Hopfian. $\endgroup$
    – YCor
    Dec 25, 2012 at 10:59
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    $\begingroup$ @HW here's, I think, a non-hopfian group $G$ with an index 2 hopfian group. Let $S=sym(5)$, $A=alt(5)$, and G the direct sum of countably copies of $S$ (clearly not hopfian). Let $H$ be its index 2 subgroup given as kernel of the map sum of the signature maps of the factors. I think that $H$ is hopfian. The point is to check that any proper quotient of $H$ with trivial center is isomorphic to a direct sum of copies of $S$. $\endgroup$
    – YCor
    Dec 25, 2012 at 15:28
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    $\begingroup$ Yves - nice work! $\endgroup$
    – HJRW
    Dec 25, 2012 at 22:09

1 Answer 1


Here are the details for my comment with a "direct proof" (by reduction to the free group case) of Hopfian property for groups $G=H_1*...*H_k$, where each $H_i$ is a finite group, say, $Z_2$ in OP's question. (Note that I did not claim that such reduction holds in general, Yves constructed a nice counter-example to such a claim, see his comment above.)

Let $f: G\to G$ be an epimorphism. Every finite order element of $G$ is conjugate to one of the subgroups $H_i$. I claim that kernel of $f$ is torsion-free: Otherwise consider the projection $$ \bar{f}: \bar{G}=H_1\times ...\times H_k \to \bar{G}. $$ Then $\bar{f}$ is surjective and not injective homomorphism of finite groups, which is impossible.

Let $F\subset G$ be a free subgroup of finite index, let $F'=f^{-1}(F)$. By the above observation, $F'$ is also torsion-free, hence, free. Clearly, $i=|G:F|=|G:F'|$. Therefore, $$ \chi(F)=\chi(F')= i\chi(G). $$ Here I am using the virtual Euler characteristic $\chi(K)$ of groups $K$, which is also the Euler characteristic of the suitable hyperbolic orbifold $O$ so that $K=\pi_1(O)$.

Since both $F, F'$ are free groups, $\chi(F)=1-r$, $\chi(F')=1-r'$, where $r$, $r'$ are ranks of $F$ and $F'$ respectively. Thus, $f$ restricts to an epimorphism $f': F'\cong F\to F$. This epimorphism has to be injective since $F$ is Hopfian. Clearly, $Ker(f)=Ker(f')$, so injectivity of $f'$ implies injectivity of $f$. qed

I'd say that this argument is not any better than the argument which uses residual finiteness of $G$. There is also an alternative argument proving Hopfian property for $G$ which uses group actions on simplicial trees directly, without using Hopfian property for ree groups.

  • $\begingroup$ +1 Thanks for this answer. I agree it's probably no better than the residual finiteness argument, since you have to bring in Euler characteristics, but it is very interesting nonetheless. If no simpler proof is produced (there very well may not be one at all) I'll accept this answer. $\endgroup$ Dec 26, 2012 at 6:34
  • $\begingroup$ @Misha: I am wondering about a proof that f.g. free groups are Hopfian using actions on trees, you mention it at the end of your answer. What is the idea behind it? $\endgroup$ Nov 15, 2023 at 10:29
  • $\begingroup$ @AshotMinasyan: Unfortunately, 11 years later I do not remember what I had in mind. Maybe it was Sela's proof that hyperbolic groups are Hopfian... $\endgroup$
    – Misha
    Dec 7, 2023 at 21:12

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