This is a follow-up to an earlier post, although my question here is self-contained.
Let $G$ be a group, $S$ a subset of $G$. Denote $\langle S \rangle _G$ the normally-generated subgroup generated by $S$, i.e. the subgroup generated by $\{ g^{-1}sg : s \in S, g \in G\}$. Also, denote $Q=G/\langle S \rangle _G$.
Now let's name some maps:
- Let $\varphi_G$ be the quotient map from $G$ to $G_{\mathrm{ab}}$
- Let $q_S$ the quotient map from $G$ to $Q$
- Let $\varphi_Q$ be the quotient map from $Q$ to $Q_{\mathrm{ab}}$
- Let $p_S$ be the quotient map from $G_{\mathrm{ab}}$ to $G_{\mathrm{ab}}/ \langle \{\varphi_G(x):x \in S \} \rangle $.
Per the comments in the linked post:
We have $Q_{\mathrm{ab}}=(G/\langle S \rangle _G)_{\mathrm{ab}}=(G/\langle S \rangle _G)/[G/\langle S \rangle _G:G/\langle S \rangle _G] \cong G/([G:G]\langle S \rangle _G)$. This is equal to (isomorphic to?) $(G/[G:G])/\langle \{ \varphi_G(x) : x \in S\}\rangle = G_{\mathrm{ab}}/ \langle \{ \varphi_G(x) : x \in S \} \rangle $.
In words, we have that the codomain of $p_S \circ \varphi_G$ is isomorphic to the codomain of $\varphi_Q \circ q_S$.
My question is whether the diagram with $G, Q, G_{\mathrm{ab}}, Q_{\mathrm{ab}}$ and the above maps is commutative. That is, we do we have $p_S \circ \varphi_G = \varphi_Q \circ q_S$? (I'm not certain my question is well-defined, since I've only shown the codomains are isomorphic--does it even make sense then to ask whether those two compositions are 'equal' if the codomains are not exactly equal?)