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Could someone please indicate the proof of the following fact (I believe strongly, but I am not sure, that this is true). Let $F$ be a free group of finite rank, $N<F$ a normal subgroup and $G=F/N$. Let $X$ be the Cayley graph of $G$. Then $\pi_1(X)$ is isomorphic to $N$.

I can prove the existence of an epimorphism $N\to \pi_1(X)$, but cannot prove that it is injective. (The homomorphism associates to a word in $N$ the class of the loop obtained by lifting the word to $X$.)

Also, I have the following related question: Suppose that $M<N<F$ (proper inclusions), with $F,N$ as above and $M$ normal in $F$. Is it possible that $F/M\simeq F/N$?

Thank you very much for the help.

rgnrmllbrg
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2 Answers2

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The answer for the second question is "Yes": not all groups (even finitely generated) are Hopfian.

For the first question, it is also obviously "Yes": the graph of $F$, which is contractible, covers that of $G$.

Alex Degtyarev
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  • Thank for your answer. Still regarding the first question, I do not understand why it is obvious that $\pi_1(X)\simeq N$ follows from the fact that $F$ covers $X$: $F$ covers the Cayley graph of any group generated by (rank $F$) elements. Perhaps it is obvious to you that the group of deck transformations of $X_F\to X_G$ is isomorphic to $N$? I cannot see this. Could you clarify? –  Mar 14 '14 at 00:07
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    The Cayley graph of $G$ is the quotient of the Cayley tree of F by the fixed point free (and hence properly discontinuous) action of $N$. Thus the group of deck transformations is $N$ by standard covering space theory. – Benjamin Steinberg Mar 14 '14 at 01:25
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Actually, the fundamental group of a graph is always a free group, so it can not be $F/N$. To get a space with $\pi_1X=F/N$ you will have to fill in 2-cells whose boundaries correspond to elements of $N$.

user39082
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