Did anyone find a proof of this theorem? I can't find it on the Internet. The theorem is :
Let $X$ be a set and let $R$ be a set of reduced words on $X$. Assume that a group $G$ has the presentation $\langle X \mid R \rangle$. If $H$ is any group generated by $X$ and satisfies the relations of $G$, i.e., $w=1$ in $H$ for all $w \in R$, then there is a surjective group homomorphism from $G$ to $H$.
Thanks in advance.
The result is essentially a consequence of the universal property of free groups:
Let $G= \langle X \mid R \rangle$ and let $H$ be any group generated by $X$ satisfying $w=1$ for any $w \in R$. Let $F$ denote the free group over $X$. Clearly, there exists an epimorphism $\varphi : F \twoheadrightarrow H$ such that $\varphi(X)=X$. Now, let $g \in \langle \langle R \rangle \rangle \leq F$. Then $g$ can be written as a product
$$g = w_1^{r_1} \cdots w_n^{r_n}$$
where $w_i \in R$ and $r_i \in F$. Now, $\varphi(g) = \varphi(w_1)^{r_1} \cdots \varphi(w_n)^{r_n} = w_1^{r_1} \cdots w_n^{r_n} = 1$ in $H$. Therefore,
$$\overline{\varphi} : \left\{ \begin{array}{ccc} G= F/ \langle \langle R \rangle \rangle & \to & H \\ \overline{g} & \mapsto & g \end{array} \right.$$
is a well-defined epimorphism.
For a reference, the result probably can be found in Johnson's book Presentations of groups.
