Exercise 15, § 7

Let G be a group with presentation $(\mathbf{t}, \mathbf{r})$, with $\mathbf{t} = (t_i)_{i \in I}$ and $\mathbf{r} = (r_j)_{j \in J}$; denote by F((T_i)_{i \in I}) the free group F(I), and put $\mathbf{T} = (T_i)$. Let on the other hand $\mathbf{x} = (x_\alpha)_{\alpha \in A}$ be a generating family of G; put $F(A) = F((X_\alpha)_{\alpha \in A}), \mathbf{X} = (X_\alpha)$, and denote by $R_\mathbf{x}$ the set of relators of $\mathbf{x}$; it is a normal subgroup of F(A). For every $i \in I$, let $\varphi_i(\mathbf{X})$ be an element of F(A) such that $t_i = \varphi_i(\mathbf{x})$ in G; for every $\alpha \in A$, let $\psi_\alpha(\mathbf{T})$ be an element of F(I) such that $x_\alpha = \psi_\alpha(\mathbf{t})$ in G. Now prove that $R_\mathbf{x}$ is the normal subgroup of F(A) generated by the elements $X_\alpha^{-1} \psi_\alpha(\varphi_i(\mathbf{X})_{i \in I})$ for $\alpha \in A$, and $r_j(\varphi_i(\mathbf{X})_{i \in I})$ for $j \in J$. (If $R'_\mathbf{x}$ denotes the normal subgroup of F(A) generated by the elements in question, one has $R'_\mathbf{x} \subset R_\mathbf{x}$; on the other hand, one will show that the homomorphism $F(I) \to F(A)$ defined by the $\varphi_i$ gives, by passing to the quotient, a homomorphism $G \to F(A)/R'_\mathbf{x}$ inverse to the canonical homomorphism $F(A)/R'_\mathbf{x} \to F(A)/R_\mathbf{x} = G$.)