## Group Theory I.4 (Conjugate Permutations)

Let $\sigma, \tau \in S_n$ be any permutations. The element $\tau \sigma \tau^{-1} \in S_n$ is called a conjugate of $\sigma$. [ Note: for brevity of notation, we’ve omitted the composition symbol $\circ$. ]

Let us give a more descriptive formulation of this. We trace the elements 1, 2, …, n through the permutation $\tau \sigma \tau^{-1}$:

• first, $\tau^{-1}$ takes $\tau(i)$ to i;
• next, $\sigma$ takes i to $\sigma(i)$;
• finally, $\tau$ takes $\sigma(i)$ to $\tau(\sigma(i))$.

In short, the conjugate permutation takes $\tau(i) \mapsto \tau(\sigma(i))$, i.e. if we write $\sigma$ as a product of disjoint cycles (see above),

$\sigma = (a_1, a_2, \ldots, a_r) (b_1, b_2, \ldots, b_s) (c_1, c_2, \ldots, c_t) \ldots$

then the conjugate permutation $\tau \sigma \tau^{-1}$ is obtained by replacing each occurring number k by $\tau(k)$:

$\tau \sigma \tau^{-1} = (\tau(a_1), \ldots, \tau(a_r)) (\tau(b_1), \ldots, \tau(b_s)) (\tau(c_1), \ldots, \tau(c_t)) \ldots$

Hence, we have the following result:

The permutation $\sigma^\prime\in S_n$ is a conjugate of $\sigma \in S_n$ if and only if they have the same cycle structure.

The astute reader may point out that we’ve only demonstrated (LHS implies RHS). The converse is really quite easy to show, as the following example should amply illustrate:

Example : Show that the following permutations of S8 are conjugate, and find a $\tau$ such that $\sigma^\prime = \tau \sigma \tau^{-1}$.

$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ 3 & 7 & 6 & 2 & 5 & 1 & 8 & 4 \end{pmatrix},$

$\sigma^\prime = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ 5 & 1 & 8 & 3 & 2 & 4 & 7 & 6 \end{pmatrix}.$

Answer : Write them as a product of disjoint cycles, thus giving (1,3,6)(2,7,8,4) and (1,5,2)(3,8,6,4) respectively. Thus, we can pick $\tau \in S_8$ which maps 1,3,6,2,7,8,4 to 1,5,2,3,8,6,4 respectively. This gives:

$\tau = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 3 & 5 & 4 & 7 & 2 & 8 & 6 \end{pmatrix}.$

Note that this is not the only answer!

Exercise : compute the number of solutions for $\tau$.