# Composite of Permutations is Permutation

Jump to navigation
Jump to search

## 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