Permutation Group is Subgroup of Symmetric Group

Theorem
Let $S$ be a set.

Let $\left({\Gamma \left({S}\right), \circ}\right)$ be the symmetric group on $S$, where $\circ$ denotes the composition operation.

Let $\left({H, \circ}\right)$ be a set of permutations of $S$ which forms a group under $\circ$.

Then $\left({H, \circ}\right)$ is a subgroup of $\left({\Gamma \left({S}\right), \circ}\right)$.

Proof
Follows directly from the definition of subgroup:

$H$ is a subset of $\Gamma \left({S}\right)$, and $\left({H, \circ}\right)$ is a group.

Hence the result.