Parity of Inverse of Permutation
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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 $9.17$