Let be any permutations. The element is called a **conjugate** of . [ Note: for brevity of notation, we’ve omitted the composition symbol . ]

Let us give a more descriptive formulation of this. We trace the elements 1, 2, …, *n* through the permutation :

- first, takes to
*i*;
- next, takes
*i* to ;
- finally, takes to .

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

then the conjugate permutation is obtained by replacing each occurring number *k* by :

Hence, we have the following result:

The permutation is a conjugate of 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* **S*_{8} are conjugate, and find a such that .

*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 which maps 1,3,6,2,7,8,4 to 1,5,2,3,8,6,4 respectively. This gives:

Note that this is not the only answer!

*Exercise : compute the number of solutions for *.

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Thanks. Going through the exercises in Charles Pinter’s A Book of Abstract Algebra and this one had me stumped.