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Finiding a representative of the 2nd conjugacy class in A_n

  1. Mar 8, 2013 #1
    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.

    Thanks in advance.
     
  2. jcsd
  3. Mar 8, 2013 #2
    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.
     
  4. Mar 9, 2013 #3

    jbunniii

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    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##.
     
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