Composite of Permutations is Permutation

Theorem

Let $f, g$ are permutations of a set $S$.


Then their composite $g \circ f$ is also a permutation of $S$.


Proof

This follows from the fact that a permutation is a bijection.

The domain and codomain are coincident.

The result follows from Composite of Bijections is Bijection.

$\blacksquare$


Sources

  • 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.3: \ 5$
  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations
  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.1$. Binary operations on a set: Example $59$
  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 25.4 \ \text{(i)}$: Some further results and examples on mappings
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