Group of Permutations is a Group
Contents |
Theorem
Let $S$ be a set.
Let $\Gamma \left({S}\right)$ denote the set of all permutations on $S$.
Then the group of permutations on $S$ $\left({\Gamma \left({S}\right), \circ}\right)$ forms a group.
It is a subgroup of $\left({S^S, \circ}\right)$, where $S^S$ is the set of all mappings on $S$.
Proof 1
Taking the group axioms in turn:
G0: Closure
A Composite of Permutations on $S$ is itself a permutation on $S$, and thus $\left({\Gamma \left({S}\right), \circ}\right)$ is closed.
$\Box$
G1: Associativity
From Set of All Mappings is a Monoid, we already have that $\left({\Gamma \left({S}\right), \circ}\right)$ is associative.
$\Box$
G2: Identity
Also from Set of All Mappings is a Monoid, we already have that $\left({\Gamma \left({S}\right), \circ}\right)$ has an identity, that is, the identity mapping.
$\Box$
G3: Inverses
By Inverse of Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
$\Box$
As a permutation is a mapping, it follows that $\Gamma \left({S}\right) \subseteq S^S$.
Thus by definition $\left({\Gamma \left({S}\right), \circ}\right)$ is a subgroup of $\left({S^S, \circ}\right)$.
$\blacksquare$
Proof 2
A direct application of Invertible Mappings form Group of Permutations.
$\blacksquare$
Sources
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 7$: 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): $\S 26 \alpha$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 76$
