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.