In one of our lectures for a course, my professor assumed something which is not obvious to me. Or at least, the proof is not obvious to me.
Let $G$ be a group, $S$ a subset. Denote $\langle S \rangle ^G$ the subgroup normally generated by $S$, i.e. $\langle S \rangle ^ G =$ the subgroup generated by $ \{ g^{-1}sg : g \in G, s \in S\}$. If we quotient $G$ by $\langle S \rangle ^G$ and then abelianize, do we get (up to isomorphism) the same group obtained by abelianizing $G$ and then quotienting by $\langle \{ \varphi_G(x): x \in S\} \rangle$, where $\varphi_G:G \to G_{\mathrm{ab}}$ is the natural homomorphism?
I wish I could give some commentary but this is my first time handling most of these things. It would help to label everything: let $\varphi_G:G \to G_{ab}$ be the natural homomorphism, $q:G \to G/\langle S \rangle^G$ the quotient map, $p:G_{ab} \to G_{ab}/\langle \varphi_G(S) \rangle$ the quotient map, and $\varphi_{G/\langle S \rangle^G} :G/\langle S \rangle^G \to (G / \langle S \rangle^G)_{ab}$ the natural homomorphism. Then my question is, how do we know $G_{ab}/\langle \varphi_G(S) \rangle$ is $(G / \langle S \rangle ^G)_{ab}$?