An immediate consequence of Corollary 9.3 and the First Isomorphism Theorem is that any group $G$ is isomorphic to a quotient group $F/N$, where $G =\langle X\rangle$, $F$ is the free group on $X$ and $N$ is the kernel of the epimorphism $F\to G$ of Corollary 9.3. Therefore, in order to describe $G$ up to isomorphism we need only specify $X$, $F$, and $N$. But $F$ is determined up to isomorphism by $X$ (Theorem 7.8) and $N$ is determined by any subset that generates it as a subgroup of $F$. Now if $w=x_1^{\delta_1}\cdots x_n^{\delta_n}$ is a generator of $N$, then under the epimorphism $F\to G$, $w\mapsto x_1^{\delta_1}\cdots x_n^{\delta_n} = e \in G$. The equation $ x_1^{\delta_1}\cdots x_n^{\delta_n} = e$ in $G$ is called a relation on the generators $x_i$. Clearly a given group G may be completely described by specifying a set $X$ of generators of $G$ and a suitable set $R$ of relations on these generators. This description is not unique since there are many possible choices of both $X$ and $R$ for a given group $G$ (see Exercises 6 and 9).
Question: I don’t exactly get why we need $R$ to specify $G$. Isn’t $G$ completely described by $X$? I mean $G=\langle X\rangle$.
Sentence “Now if $w=x_1^{\delta_1}\cdots x_n^{\delta_n}$ is a generator of $N$, then under the epimorphism $F\to G$, $w\mapsto x_1^{\delta_1}\cdots x_n^{\delta_n} = e \in G$” is bit convoluted, in my opinion. It simply say, if $w\in N$, then $w=e$. So $R=N$.
