Parity of Inverse of Permutation

Theorem

Let $S_n$ denote the symmetric group on $n$ letters.

Then:

$\forall \pi \in S_n: \map \sgn \pi = \map \sgn {\pi^{-1} }$


Proof

From Parity Function is Homomorphism:

$\map \sgn {I_{S_n} } = 1$

Thus:

$\pi \pi^{-1} = I_{S_n} \implies \map \sgn \pi \, \map \sgn {\pi^{-1} } = 1$

The result follows immediately.

$\blacksquare$


Sources

  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $9$: Permutations: Corollary $9.17$
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