Permutation Group is Subgroup of Symmetric Group

Theorem

Let $S$ be a set.

Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$, where $\circ$ denotes the composition operation.

Let $\struct {H, \circ}$ be a set of permutations of $S$ which forms a group under $\circ$.


Then $\struct {H, \circ}$ is a subgroup of $\struct {\map \Gamma S, \circ}$.


Proof

Follows directly from the definition of subgroup:

$H$ is a subset of $\map \Gamma S$, and $\struct {H, \circ}$ is a group.

Hence the result.

$\blacksquare$


Sources

  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 36$: Subgroups: Simple illustrations: $(3)$
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