I was reading this discussion to know whether a conjugacy class of $S_n$ splits or not when we go down to $A_n$. With this in mind when I read this wiki, I am confused when looking at the table called "particular cases". Indeed if we consider the second row i.e. we consider a permutation with decomposition 1+1+1 then we have 3 1-cycles so according to the criterion it should not split (since there are 2 cycles of same length). Also I was wondering why when the class splits, it splits only in two classes and not 3,4,5,... And why if it splits in 2 classes the respective classes cardinality is divided by 2 (it could be that more elements of the original conjugacy class go into one of the splitting class than the other no?).
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Try my answer here: https://math.stackexchange.com/questions/3415844/a-n-and-the-conjugacy-classes/3773629#3773629 – David A. Craven Aug 01 '20 at 11:54
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Why $|C_{S_n}(g):C_{A_n}(g)|\in{1,2}$? – roi_saumon Aug 01 '20 at 15:30
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Second isomorphism theorem as $C_H(X)=C_G(x)\cap H$. – David A. Craven Aug 01 '20 at 15:31
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$C_G(x)/(C_G(x)\cap H)\cong C_G(x)H/H$ but why is this of cardinality 2? – roi_saumon Aug 01 '20 at 15:45
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By the way, is $g^x$ a notation for $xgx^{-1}$? – roi_saumon Aug 01 '20 at 15:49
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$C_G(x)H$ is either $G$ or $H$ since $|G:H|=2$. And $g^x$ is either $x^{-1}gx$ or $xgx^{-1}$, depending on whom you talk to. – David A. Craven Aug 01 '20 at 16:05
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Oh, right! But still, concerning my question, I don't get why the permutation for which the conjugacy class "is" 1+1+1 splits. According to the criterion it should not since it has 2 (even 3) cycles of same length. – roi_saumon Aug 01 '20 at 16:24
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1It doesn't. You aren't reading the table properly. – David A. Craven Aug 01 '20 at 17:11