Exercise on Generators and Relations from Michael Artin's book

Question

The question is:

Let $\phi: G \mapsto G'$ be a surjective group homomorphism. Let $S$ be a subset of $G$ whose image under $\phi$(S) generates $G$', and let $T$ be a set of generators of $\ker\phi$. Prove that $S \cup T$ generates $G$.

Firstly, I tried proving that any element $g \in G$ belongs to $\langle S \cup T \rangle$. The farthest I went like this was finding that $$\phi^{-1}\left(G'\right) = \{g \in G : \phi(g) \in G'\} = S \cup T \cup L = G$$ with an undefined set $L$.

I then tried using some of the theorems, such as the First Isomorphism Theorem and the fact that every group is isomorphic to the quotient of a free group. I got the following, immediate results: $$ {G\over <T>} \cong G', {F(T)\over \ker f_1} \cong \ker\phi, {F(S \cup T)\over \ker f_2} \cong <S \cup T>, {F(\phi(S))\over \ker f_3} \cong G'$$

Answer 1

Take any $g \in G$, then $\phi(g) \in G'=\langle \phi(S)\rangle$ by assumption. Thus we can write $\phi(g)=\phi(s)$ with $s \in \langle S\rangle$. This implies that $\phi(gs^{-1})=1$, so $gs^{-1} \in \ker(\phi)$ and we have $gs^{-1} \in \langle T\rangle$. This means that $g \in \langle S\cup T \rangle$.