How many conjugates does the permutation (123) have in the group S<sub>3</sub> of all permutations on 3 letters? Give brief reasons for your answers.
Answers were provided for this question, but after going carefully through it, I still don't know what's going on at one point in the answer, which may prove crucial to my understanding of the problem (where I've put question marks). Below is the solution provided:
"We can use the stabilizer-orbit relationship to see that the size of the conjugacy class of (123) is equal to the index of the stabilizer. But the stabilizer is a subgroup of S3 which contains at least <(123)> [????? why]. If it were to contain more then it must be all of S3 (by Lagrange's theorem) and so contain, for example, (12). But (12)(123)(12)-1 = (132) and so the centralizer is exactly <(123)> [????]. Since it has order 3, it also has index 2 and so (123) has 2 conjugates."
Originally Posted by davido
I see now, thanks so much. I didn't take much notice of "any element in any group commutes with itself and, in fact, with any of its powers", especially about the powers.