# Finiding a representative of the 2nd conjugacy class in A_n

1. Mar 8, 2013

### skojoian

Hey everyone, im trying to figure this out, but no luck. Help would be much appreciated. So i am considering a 5-cycle: pi = (12345) in S_5. Its centralizer in S_5 is just <(pi)>, which is contained in A5. Therefore, when we restrict our main group to A_5, the conjugacy class that contained ALL permutations of cycle structure 5^1 gets split into two conjugacy classes. Now, a representative of the first one is obviously (12345) itself. Which brings me to my question:

Q: How to find the rep. of the 2nd conjugacy class?

I asked a group theorist, and here is his response:
(15)(24)(12345)(15)(24)=(54321)=[(12345)][/-1]. It follows that pi and pi^2 = (13524) are the 2 reps. In other words, (12345) and (13524) are NOT conjugate.

I am having trouble understanding exactly how this implies they are not conjugate.

2. Mar 8, 2013

### skojoian

yes, jbuni, but i am restricting my group from S_5 to A_5 and the permutations you listed are odd, hence, not in A_5. Its all in the description of the problem i wrote above.

3. Mar 9, 2013

### jbunniii

I'm not sure what your group theorist was getting at, but it's pretty straightforward to find a representative of the second conjugacy class.

Put $r = (12345)$ and $s = (21345) = (13452)$. Then if $g = (12)$, we have $g r g^{-1} = s$. Moreover, any $h \in S_5$ satisfies $hrh^{-1} = s$ if and only if $h = gz$ for some $z$ in the centralizer of $r$. But every such $z$ is in $A_5$, so $h = gz = (12)z \not\in A_5$. This shows that $r$ and $s$ are not conjugate in $A_5$.