Symmetric Group is Group

Theorem
The symmetric group on $n$ letters $\left({S_n, \circ}\right)$ is isomorphic to the Group of Permutations of the $n\,$ elements of any set $T$ whose cardinality is $n$.

That is:
 * $\forall T \subseteq \mathbb U, \left|{T}\right| = n: \left({S_n, \circ}\right) \cong \left({\Gamma \left({T}\right), \circ}\right)$

Proof
The fact that $\left({S_n, \circ}\right)$ is a group follows directly from Group of Permutations.

By definition of cardinality, as $\left|{T}\right| = n$ we can find a bijection between $T$ and $\N_n$.

From Number of Permutations, it is immediate that $\left|{\left({\Gamma \left({T}\right), \circ}\right)}\right| = n! = \left|{\left({S_n, \circ}\right)}\right|$.

Again, we can find a bijection $\phi$ between $\left({\Gamma \left({T}\right), \circ}\right)$ and $\left({S_n, \circ}\right)$.

The result follows directly from the Transplanting Theorem.