I'm doing Exercise 10 in textbook Algebra by Saunders MacLane and Garrett Birkhoff. Could you please verify if my attempt is fine or contains logical mistakes?
Let $\phi:G \to H$ be a group homomorphism such that $N = \operatorname{ker} \phi$ and $S$ is a subgroup of $G$. The induced map $\phi[\cdot]$ is defined as $\phi[S] = \{\phi(x) \mid x \in S\}$. Then $$[G: S]=[\phi [G]: \phi [S]] \cdot[N: S \cap N]$$
My attempt: First, we need a lemma:
If $N \le S \le G$, then $x \in S \iff \phi(x) \in \phi[S]$.
Notice that $$[G: S]=[\phi [G] : \phi [S]] \cdot [N : S \cap N] \iff \frac{|G|}{|S|} = \frac{|\phi [G] |}{|\phi [S]|} \cdot \frac{|N|}{|S \cap N|}$$
By second isomorphism theorem on groups, we have ${|SN|} / {|N|} = {|S|} / {|S \cap N|}$. Moreover, $\phi [SN] = \phi [S]$. Then our job boils down to proving ${|G|} / {|SN|} = {|\phi [G] |} / {|\phi [SN] |}$. Consider the map $$\psi: G/SN \to \phi [G] / \phi [SN], \quad gSN \mapsto \phi(g) \phi [SN]$$
It's sufficient to show that $\psi$ is bijective. Let $x,y \in G$ such that $xSN = ySN$. We have $xSN = ySN$ iff $x^{-1}y \in SN$ iff $(\star)$ $\phi(x^{-1}y ) \in \phi [SN]$ iff $\phi(x)^{-1} \phi(y) \in \phi [SN]$. Hence $\psi$ is both well-defined and injective. Clearly, $\phi$ is surjective. This completes the proof.
$(\star)$ This is due to our lemma and the fact that $N \le SN$.
I would use the first isomorphism theorem to get that $\phi(G)\cong G/N$ And similarly we have that $\phi(S)\cong S/(N\cap S)$ From here it should be a breeze since and isomorphism is a bijection so $|\phi(G)|= |G/N|$ and $|\phi(S)|= |S/(N\cap S)|$ Divide and you are done
Edit: $\phi(S)\cong S/(N\cap S)$ follows from the restriction of $\phi$ to $S$