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I have been watching the linked video on character tables of $S_4$ and $A_4$. I understand the process used in $S_4$, but I am having difficulty with that of $A_4$.

The conjugacy classes of $S_4$ are given in the video by $\{id, [(12)], [(123)], [(12)(34)], [(1234)]\}$, this I understand.

However when moving to $A_4$ the classes are given by $\{id, [(123)], [(132)], [(12)(34)]\}$. Why are the three cycles now represented by two conjugacy classes?

I do not see how this result is arrived at.

Video: https://www.youtube.com/watch?v=POh_-mdBHVw

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  • $\begingroup$ Did you delete a question of yours about $\mathbb C[A_4]$ in the past 10 hours? I had been intending to answer it. $\endgroup$ – rschwieb Feb 14 '17 at 12:12
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In $S_4$, the two $3$-cycles are conjugate:

$(2,3)(1,3,2)(2,3)=(1,2,3)$.

But in $A_4$, you have less things to conjugate with, so pairs that used to be conjugate might not be conjugate anymore. Like here, you cannot use $(2,3)$ to conjugate again, because $(2,3)\notin A_4$.

I do not see how this result is arrived at.

Surely you can just compute the conjugacy class of each element on your own, if it seems unclear.

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  • $\begingroup$ Thanks a lot, that made it clear. $\endgroup$ – ushham Feb 15 '17 at 10:10

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