Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)?

share|improve this question
1  
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. –  Misha Dec 25 '12 at 6:22
    
@Misha Ah, I see. I was stuck viewing $G$ as a finitely presented group and failed to consider it as $\mathbb Z_2*\mathbb Z_2*\mathbb Z_2$. –  Alex Becker Dec 25 '12 at 6:46
    
Incidentally, hyperbolic groups are known to be Hopfian, even though they are not known to be residually finite in general. –  Ian Agol Dec 25 '12 at 7:36
    
Misha - are you claiming that a virtually Hopfian group is Hopfian? This isn't completely obvious to me (perhaps I'm missing something obvious). –  HJRW Dec 25 '12 at 8:02
3  
@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. –  Yves Cornulier Dec 25 '12 at 10:59
show 5 more comments

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
    
+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. –  Alex Becker Dec 26 '12 at 6:34
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.