Fundamental group of a Cayley graph

Question

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.

Answer 1

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 $G$.

Answer 2

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 $N$.