Second isomorphism theorem

Statement

Let $G$ be a group, let $S$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$. Then $SN$ is a subgroup of $G$, $N$ is normal in $SN$, and $S \cap N$ is normal in $S$. Moreover, the map $s(S \cap N) \mapsto sN$ is an isomorphism from $S/(S \cap N)$ to $SN/N$.

Remarks

The proof applies the first isomorphism theorem to the homomorphism $S \to SN/N$, $s \mapsto sN$, which is surjective with kernel $S \cap N$. For finite groups, it gives $|SN| = |S| \, |N| / |S \cap N|$. It is enough that $S$ normalizes $N$, that is, $sNs^{-1} = N$ for each $s \in S$.

Explanations

L1 human

Picture a diamond with $SN$ at the top, $S$ and $N$ in the middle, and $S \cap N$ at the bottom. The theorem says that the two slanted sides opposite each other give the same quotient group. For integers, with $S = 4\mathbb{Z}$ and $N = 6\mathbb{Z}$, it gives $4\mathbb{Z}/12\mathbb{Z} \cong 2\mathbb{Z}/6\mathbb{Z}$, both cyclic of order $3$.