Set of Invertible Mappings forms Symmetric Group

Theorem

Let $S$ be a set.

Let $\GG$ be the set of all invertible mappings from $S$ to $S$.

Then $\struct {\GG, \circ}$ is the symmetric group on $S$.

Proof

Let $\struct {S^S, \circ}$ be the algebraic structure formed from the set of all mappings from $S$ to itself.

From Set of all Self-Maps under Composition forms Monoid, $\struct {S^S, \circ}$ is a monoid.

By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.

By Bijection iff Inverse is Bijection, it follows that all the invertible elements of $S^S$ are exactly the permutations on $S$.

The result follows from Invertible Elements of Monoid form Subgroup of Cancellable Elements.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets