Smallest Normal Subgroup containing Set

Theorem

Let $S \subseteq G$ where $G$ is a group.

Then there exists a unique smallest normal subgroup of $G$ which contains $S$.

Proof

Let $\Bbb S$ be the set of all normal subgroups of $G$ that contain $S$.

Since $S \subseteq G \lhd G$, $\Bbb S \ne \O$.

Let $N = \bigcap \Bbb S$, that is, the intersection of all elements of $\Bbb S$.

By Intersection of Normal Subgroups is Normal, $N \lhd G$.

By the definition of intersection, $S \subseteq N$.

By the method of construction, $N$ is the smallest such subgroup.

By the definition of "smallest", $N$ is unique.

$\blacksquare$