Finiding a representative of the 2nd conjugacy class in A_n

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In summary, the conversation discusses the splitting of a conjugacy class in a group, with the question being how to find a representative of the second conjugacy class. The group theorist suggests using ##r = (12345)## and ##s = (21345) = (13452)##, and demonstrates that these two permutations are not conjugate in ##A_5## by showing that any element that conjugates ##r## to ##s## is not in ##A_5##.
  • #1
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Hey everyone, I am 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.
 
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  • #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.
 
  • #3
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##.
 

1. How do I find a representative of the 2nd conjugacy class in An?

To find a representative of the 2nd conjugacy class in An, first determine the cycle structure of the elements in An. Then, look for elements that have the same cycle structure but are not conjugate to each other. This will give you the representatives for the 2nd conjugacy class.

2. What is the significance of finding a representative of the 2nd conjugacy class in An?

Finding a representative of the 2nd conjugacy class in An is important because it allows us to understand the structure and behavior of the group An. It also helps in solving certain problems and making connections with other mathematical concepts.

3. Can there be more than one representative of the 2nd conjugacy class in An?

Yes, there can be multiple representatives of the 2nd conjugacy class in An. This is because elements with the same cycle structure may not necessarily be conjugate to each other in the group An.

4. How does the size of An affect the process of finding a representative of the 2nd conjugacy class?

The size of An does not have a direct impact on the process of finding a representative of the 2nd conjugacy class. The process remains the same regardless of the size of the group. However, the number of representatives may vary depending on the value of n.

5. Are there any specific techniques or algorithms for finding a representative of the 2nd conjugacy class in An?

Yes, there are certain techniques and algorithms that can be used to find a representative of the 2nd conjugacy class in An. These include the cycle decomposition method, the symmetric polynomials method, and the Young tableaux method. However, the most efficient method may vary depending on the specific problem and the value of n.

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