Suppose we have a group $A_n$ for some $n$ (maybe take $A_5$ as an example).

We find the conjugacy classes of $S_n$ which are determined by cycle type. Then we use the splitting criterion ,http://groupprops.subwiki.org/wiki/Splitting_criterion_for_conjugacy_classes_in_the_alternating_group, as we find a conjugacy class that splits, eg that represented by $(12345)$ in $S_5$.

Is there a way of computing/spotting the two conjugacy classes. One will be $(12345)$ but is there a way of getting the other preferably without doing any calculations.

Looking for general method.

  • $\begingroup$ General comment. For $H\leq G$ finite group, and $x\in G$ we have $x^G\cap H=\bigcup_{i\leq|G:H|}x_i^H$. I'm not aware of a general method, apart from the 'brute force' one: checking if there are two permutations in $x^{S_n}$ that are not $A_n$-conjugate. $\endgroup$
    – rafforaffo
    May 1 '15 at 8:32
  • $\begingroup$ I'm not certain about this, but I suspect $x$ and $x^{-1}$ will be in different conjugacy classes in the situation you describe. $\endgroup$ May 1 '15 at 9:14
  • $\begingroup$ Thats the sort of thing I was thinking but I cannot see any justification $\endgroup$
    – Trajan
    May 1 '15 at 9:21
  • 1
    $\begingroup$ @GerryMyerson: $(2,5)(3,4)$ conjugates $x=(1,2,3,4,5)$ to $x^{-1}$. But $x^2$ is in the other class in $S_5$. $\endgroup$
    – Derek Holt
    May 1 '15 at 9:22
  • $\begingroup$ So GerryMyerson's idea is false $\endgroup$
    – Trajan
    May 1 '15 at 9:52

The class of $g \in S_n$ splits into two classes in $A_n$ if and only if all cycles of $g$ have odd length and their lengths are all distinct (this includes cycles of length $1$).

In a class that splits, you get an element in the other class by conjugating by any odd permutation, such as $(1,2)$.


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