Parity of Conjugate of Permutation

Theorem

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

Then:

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

where $\map \sgn \pi$ is the sign of $\pi$.

Proof

As $\map \sgn \pi = \pm 1$ for any $\pi \in S_n$, we can apply the laws of commutativity and associativity:

\(\ds \map \sgn \pi \, \map \sgn \rho\)

\(=\)

\(\ds \map \sgn \rho \, \map \sgn \pi\)

\(\ds \leadsto \ \ \)

\(\ds \map \sgn \pi \, \map \sgn \rho \, \map \sgn {\pi^{-1} }\)

\(=\)

\(\ds \map \sgn \rho \, \map \sgn \pi \, \map \sgn {\pi^{-1} }\)

\(\ds \leadsto \ \ \)

\(\ds \map \sgn {\pi \rho \pi^{-1} }\)

\(=\)

\(\ds \map \sgn \rho\)

Parity Function is Homomorphism

$\blacksquare$

Sources

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