Cycle Decomposition of Conjugate

Theorem

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

Let $\pi, \rho \in S_n$.


The cycle decomposition of the permutation $\pi \rho \pi^{-1}$ can be obtained from that of $\rho$ by replacing each $i$ in the cycle decomposition of $\rho$ with $\map \pi i$.


Proof

Consider the effect of $\pi \rho \pi^{-1}$ on $\map \pi i$:

$\map {\pi \rho \pi^{-1} } {\map \pi i} = \map \pi {\map \rho i}$


That is:

$\pi \rho \pi^{-1}$ maps $\map \pi i$ to $\map \pi {\map \rho i}$

In the cycle decomposition of $\pi \rho \pi^{-1}$, $\map \pi i$ lies to the left of $\map \pi {\map \rho i}$, whereas in the cycle decomposition of $\rho$, $i$ lies to the left of $\map \rho i$.

$\blacksquare$


Sources

  • 1968: Ian D. Macdonald: The Theory of Groups ... (previous) ... (next): Appendix: Elementary set and number theory
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $9$: Permutations: Proposition $9.20$
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