# Fundamental Group of a Cayley Graph

1. Mar 14, 2013

### Monobrow

Suppose we have a group with presentation G = <A|R> i.e G is the quotient of the free group F(A) on A by the normal closure <<A>> of some subset A of F(A). Is it true that that fundamental group of the Cayley graph of G (with respect to the generating set A) will be isomorphic to the subgroup <<A>> in F(A)? It seems to me that this should be true (and it agrees with the facts that: a subgroup of a free group is free and the fundamental group of a graph is free) but I can't find this theorem stated anywhere....

2. Mar 14, 2013

### Bacle2

3. Mar 15, 2013

### Monobrow

Thanks for the reference, I'll take a look when I have the chance. I was thinking that I may have a method to prove what I was asking. The Cayley graph of F(A) is a tree T. Now F(A) acts freely and properly discontinuously on T, and thus any subgroup must also act freely and properly discontinuously (this is easy to show). So we have a free properly discontinuous action of <<S>> on T. Now if it were true that T/<<S>> were the Cayley graph of G then we would be done, since T is the universal cover of T/<<S>>. Again, I am convinced that this statement is true but I have haven't seen it stated in ANY reference which I find strange.

4. Mar 15, 2013

### Monobrow

I've just noticed that I wrote <<A>> when I meant <<R>> in the first post and <<S>> when I meant <<R>> in my second post, sorry about that!