# [group theory] Finite presentations and relations <S|R>

1. Jun 14, 2012

### nonequilibrium

1. The problem statement, all variables and given/known data
Let G be a finitely presented group. Suppose we have a finite generating set S. Prove that there is a finite set of relations $R \subset F_S$ such that <S|R> is a presentation of G.

2. Relevant equations
NA

3. The attempt at a solution
I don't know how to do this.

I think I can prove it for the case of G a free group:

So suppose S is a finite generating set of $F_{T}$ with the latter a finitely presented group (i.e. $T$ finite). Write $S=\{ s_1, \cdots, s_m \}$ and $T= \{t_1, \cdots, t_n \}$. Now every element in S is in effect a combination of elements in T, this we indicate with e.g. $s_1(t_1,\cdots,t_n)$. However, since S is a generating set, every element in T can also be written as a combination of elements in S, so we can write $s_1(t_1(s_1,\cdots,s_m),t_2(\cdots),\cdots,t_n(s_1,\cdots,s_m))$. It is clear that in $F_T$ we have $s_1^{-1} \cdot s_1(t_1(s_1,\cdots,s_m),t_2(\cdots),\cdots,t_n(s_1,\cdots,s_m)) = 1$. Analogously for $s_2,\cdots, s_m$. We define the set of relations R as these m words in $F_S$.

If $N_R$ is the normal closure of R in $F_S$, we need to prove that $F_T \cong \langle S | R \rangle$ or in other words that the kernel of $\pi: F_S \to F_T$ is $\textrm{Ker}(\pi) = N_R$. By the method of construction it is rather clear that $N_R \subset \textrm{Ker}(\pi)$. I'm not sure how to prove the other inclusion.

Also I've tried to use this result for the case of a free group for the case of a general group G, but I can't seem to make it work...