Existence of Unique Subgroup Generated by Subset

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $\varnothing \subset S \subseteq G$.

Let $\left({H, \circ}\right)$ be the subgroup generated by $S$.

Then $H = \left\langle {S} \right\rangle$ exists and is unique.

Also, $\left({H, \circ}\right)$ is the intersection of all of the subgroups of $G$ which contain the set $S$:


 * $\displaystyle \left\langle {S} \right\rangle = \bigcap_i {H_i}: S \subseteq H_i \le G$.

Existence
First, we prove that such a subgroup exists.

Let $\mathbb S$ be the set of all subgroups of $G$ which contain $S$.

$\mathbb S \ne \varnothing$ because $G$ is itself a subgroup of $G$, and thus $G \in \mathbb S$.

Let $H$ be the intersection of all the elements of $\mathbb S$.

By Intersection of Subgroups, $H$ is the largest element of $\mathbb S$ contained in each element of $\mathbb S$.

Thus $H$ is a subgroup of $G$.

Since $\forall x \in \mathbb S: S \subseteq x$, we see that $S \subseteq H$, so $H \in \mathbb S$.

Smallest
Now to show that $H$ is the smallest such subgroup.

If any $K \le G: S \subseteq K$, then $K \in \mathbb S$ and therefore $H \subseteq K$.

So $H$ is the smallest subgroup of $G$ containing $S$.

Uniqueness
Now we show that $H$ is unique.

Suppose $\exists H_1, H_2 \in \mathbb S$ such that $H_1$ and $H_2$ were two such smallest subgroups containing $S$.

Then, by the definition of "smallest", each would be equal in size.

If one is not a subset of the other, then their intersection (by definition containing $S$) would be a smaller subgroup and hence neither $H_1$ nor $H_2$ would be the smallest.

Hence one must be the subset of the other, and by Equality of Sets, that means they must be the same set.

So the smallest subgroup, whose existence we have proved above, is unique.