I'm trying to infer the second isomorphism theorem on groups from the first one. Could you please verify if my attempt is fine or contains logical mistakes?
Let $G$ be a group, $S \le G$, and $N \trianglelefteq G$. Then $(S N) / N \cong S /(S \cap N)$.
My attempt:
Consider the map $\phi: S N \to S /(S \cap N), \quad sn \mapsto s(S \cap N)$. Let $s_1, s_2 \in S$ and $n_1,n_2 \in N$ such that $s_1 n_1 = s_2 n_2$. We have $s_1 n_1 = s_2 n_2 \iff s_1^{-1} s_2 =n_1 n_2^{-1} \implies s_1^{-1} s_2 \in S \cap N$ $\iff s_1(S \cap N) = s_2(S \cap N)$. Hence $\phi$ is well-defined. Clearly, $\phi$ is surjective.
It follows from $S \le G$ and $N \trianglelefteq G$ that $(S\cap N) \trianglelefteq S$. Hence $S /(S \cap N)$ is a group. Next, we show that $\phi$ is a homomorphism. Let $s_1, s_2 \in S$ and $n\in N$. It follows from $N \trianglelefteq G$ that $ns_2 = s_2 k$ for some $k \in N$. Hence $\phi ((s_1n)(s_2n)) = \phi (s_1 s_2kn) = (s_1s_2) (S \cap N) =$ $(s_1(S \cap N)) (s_2 (S \cap N))=\phi (s_1n) \phi (s_2n)$.
It's not hard to verify that $\operatorname{ker} \phi := \{sn \in SN \mid (s,n) \in S \times N \text{ and } s (S \cap N) = S \cap N\} = N$. By first isomorphism theorem on groups, the result then follows.
It's perfect.
Maybe it would be slightly easier to deduce the same starting from the other direction, with a map $S\to SN/N$.