5
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
8
$\begingroup$

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 $(1,2)$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.