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

[nonequilibrium]

Homework Statement

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.

Homework Equations

NA

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...

 

[mighty2000]

Thank you for your post. Your attempt at a solution for the case of a free group is a good start. To prove the general case, we need to use some properties of finitely presented groups.

First, we know that a finitely presented group G can be defined by a presentation <S|R>, where S is a finite generating set and R is a finite set of relations. This means that any element in G can be expressed as a word in the generators S, subject to the relations R. In other words, G is isomorphic to the quotient group F_S/N_R, where N_R is the normal closure of R in F_S.

Now, since S is a finite generating set for G, we can choose a finite set of elements s_1, s_2, ..., s_n \in S that generate G. This means that every element in G can be expressed as a word in these generators. Let's call this set of generators S'.

Next, we need to show that there exists a finite set of relations R' \subset F_{S'} such that <S'|R'> is a presentation of G. To do this, we can use the fact that G is isomorphic to F_S/N_R. This means that any element in G can be expressed as a word in the generators S, subject to the relations R. But since we have chosen a finite set of generators S', we can write any element in G as a word in S', subject to some set of relations R'. This is because any element in G can be expressed as a word in S, and we can always choose a subset of S that generates G.

Therefore, we have shown that there exists a finite set of relations R' \subset F_{S'} such that <S'|R'> is a presentation of G. This completes the proof.

I hope this helps. Let me know if you have any further questions or concerns.