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 $\pi \left({i}\right)$.

Proof

Consider the effect of $\pi \rho \pi^{-1}$ on $\pi \left({i}\right)$:

$\pi \rho \pi^{-1} \left({\pi \left({i}\right)}\right) = \pi \left({\rho \left({i}\right)}\right)$

That is:

$\pi \rho \pi^{-1}$ maps $\pi \left({i}\right)$ to $\pi \left({\rho \left({i}\right)}\right)$

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

$\blacksquare$

Sources

• John F. Humphreys: A Course in Group Theory (1996): $\S 9$: Proposition $9.20$