Inverse of Permutation is Permutation

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
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