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

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    $\begingroup$ Which definition of a group presentation are you using? $\endgroup$ – Shaun Apr 2 at 16:17
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    $\begingroup$ 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 $F(S)$. $\endgroup$ – Captain Lama Apr 2 at 16:19
  • $\begingroup$ @Shaun I've edited my post $\endgroup$ – warm_fish Apr 2 at 16:26
  • $\begingroup$ @CaptainLama Can you please clarify what is a lift? $\endgroup$ – warm_fish Apr 2 at 16:27
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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.)

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  • $\begingroup$ 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? $\endgroup$ – warm_fish Apr 4 at 11:21
  • $\begingroup$ * and $\phi$ is an isomorphism $\endgroup$ – warm_fish Apr 4 at 11:51
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    $\begingroup$ 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 ... $\endgroup$ – Eric Towers Apr 4 at 15:15
  • $\begingroup$ 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 $R \cup \pi^{-1}(\phi(R'))$ $\endgroup$ – warm_fish Apr 4 at 15:34

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