# Inverse of Permutation is Permutation

Jump to navigation
Jump to search

## Theorem

If $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.

## Proof

Let $f: S \to S$ is a permutation of $S$.

By definition, a permutation is a bijection such that the domain and codomain are the same set.

From Bijection iff Inverse is Bijection, it follows $f^{-1}$ is a bijection.

From the definition of inverse relation, the domain of a relation is the codomain of its inverse and vice versa.

Thus the domain and codomain of $f^{-1}$ are both $S$ and it follows that $f^{-1}$ is a permutation.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations - 1968: Ian D. Macdonald:
*The Theory of Groups*... (previous) ... (next): Appendix: Elementary set and number theory - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 25.4 \ \text{(ii)}$: Some further results and examples on mappings