# Why is $\langle S\mid R\cup R'\rangle$ a presentation for $G/N(R')$, where $G$ is a group with presentation $\langle S\mid R\rangle?$

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 $$G \cong F(S)/N(R)$$

• Which definition of a group presentation are you using? Apr 2 '20 at 16:17
• Your intuition is correct, but instead of $\langle S \,|\, R\cup R'\rangle$, you should rather write $\langle S \,|\, R\cup \widetilde{R'}\rangle$ where $\widetilde{R'}$ is a set of lifting of elements of $R'$ to $F(S)$. Apr 2 '20 at 16:19
• @Shaun I've edited my post Apr 2 '20 at 16:26
• @CaptainLama Can you please clarify what is a lift? Apr 2 '20 at 16:27

Notice that $$F(S)$$ is a group, $$N(R) \trianglelefteq F(S)$$, $$N(R \cup R') \trianglelefteq F(S)$$, and $$N(R) \subseteq N(R \cup R')$$. By the third isomorphism theorem, $$\frac{F(S)}{N(R \cup R')} \cong \frac{F(S)/N(R)}{N(R \cup R')/N(R)}$$ or, as presentations, $$\langle S \mid R \cup R' \rangle \cong \frac{\langle S \mid R \rangle}{N(R \cup R')/N(R)} \text{.}$$ Are you able to show that the only part of $$\langle S \mid R \rangle$$ that is actually sent to the identity by the quotient on the right is $$N(R')$$? (In particular, the implicit quotient in $$\langle S \mid R \rangle$$ has already sent all of $$N(R)$$ to the identity, so only a subset of $$N(R')$$ remains to be sent there.)
• Thanks, in this way I've managed to prove $G/N(R') \cong \langle S \mid R \cup \pi^{-1}(\phi(R')) \rangle$, where $\pi$ is the quotient map to $F(S)/N(R)$ and $\phi: G \rightarrow \langle R \mid S \rangle$. Is it correct? I wrote it this way to make sense of $R\cup R'$, however $R \cup \pi^{-1}(\phi(R'))$ seems unmanageable. Can it be further simplified? Apr 4 '20 at 11:21
• * and $\phi$ is an isomorphism Apr 4 '20 at 11:51
• I don't understand the trivial group, $\langle R \mid S \rangle$, as your codomain of $G$. Assuming you mean $\langle S \mid R \rangle$. Further, $\phi$ appears to be the identity; why is it here? For each $r' \in R'$, $\phi(r' \cdot N(R)) = r' \cdot N(R)$, so $$\pi^{-1}(r' \cdot N(R)) = (r' \cdot N(R)) \cdot N(R) = r' \cdot N(R) \text{.}$$ But $N(R)$ is sent to the identity by the relations listed before the union symbol, so ... Apr 4 '20 at 15:15
• Right, I made a typo, I meant $\langle S \mid R \rangle$. $\phi$ is an isomorphism between $G$ and its presentation $\langle S \mid R \rangle := F(S)/N(R)$, it is my definition of group $G$ with presentation $\langle S \mid R \rangle := F(S)/N(R)$". So $R'\subseteq G \implies \phi(R')\subseteq F(S)/N(R) \implies \pi^{-1}(\phi(R'))\subseteq F(S)$: now I would like to simplify $R \cup \pi^{-1}(\phi(R'))$ Apr 4 '20 at 15:34