Symmetric Group is a 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.

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 4.5$: Example $79$
• Seth Warner: Modern Algebra (1965): $\S 7$: Example $7.5$
• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.6$: Theorem $5$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 30$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 77$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 34 \ (4)$