Suppose $G$ is a group with presentation $\langle S\mid R\rangle $. By this I mean $G \cong F(S)/N(R) $, where $F(S)$ is the free group generated by the set $S$ and $N(R)$ is the normal closure of the subset $R\subseteq F(S)$.
Let $R'\subseteq G$ be a subset.
Why is $\langle S\mid R\cup R'\rangle $ a presentation for $G/N(R')$?
My intuition tells me that doing $G/N(R')$ is essentially requiring the satisfaction of the additional relations in $R'$, so I should obtain $\langle S\mid R\cup R'\rangle$ (if I interpret $R'$ as a set of words in $F(S)$ (?)).
I guess that starting to work with universal properties would leave me clueless about what's really happening. Can anybody supply a more precise explanation, or sketch out a proof?
Definition: let $S$ be a set and $R\subseteq F(S)$ a subset. A group $G$ is said to have presentation $\langle S\mid R\rangle$ if $G \cong F(S)/N(R)$