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I was trying to find the conjugacy classes of $A_5$. So I started by writing out all the conjugacy classes of $S_5$ in the hope that I could just restrict the set of them. The conjugacy class representatives of $S_5$ are $$e,(12),(123),(1234),(12345),(12)(34),(12)(345)$$ So I restricted this set to even permutations to get $$e,(123),(12345),(12)(34)$$ but apparently $(12345)^2$ represents a conjugacy class as well. So I then computed it as $(13524)$. I cannot see why this would be conjugacy class as well. As far as I was aware conjugacy classes in $S_5$ are determined by cycle type, so I thought have thought this holds for the restriction to $A_5$.

Now how would I have known a priori that $(12345)^2$ represents another conjugacy class or that I was missing a conjugacy class (assuming we dont know the number of irreducible representations).

Is there a better way to approach this question?

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  • $\begingroup$ A conjugacy class consists of more than one element (most of the time). That some of the classes split up for $A_5$ is because the only elements that would work for conjugating one to the other are not even. $\endgroup$ – Tobias Kildetoft Apr 29 '15 at 11:24
  • $\begingroup$ @TobiasKildetoft Apologies these are representatives $\endgroup$ – Permian Apr 29 '15 at 11:24
  • $\begingroup$ Right, but your formulation still does not really make sense. Any element represents a conjugacy class. The important thing is which of them represent the same one (you should really just sit down and do the calculation). $\endgroup$ – Tobias Kildetoft Apr 29 '15 at 11:27
  • $\begingroup$ The classes are the identity, the 20 3-cycles, the 15 double transpositions, 12 of the 5-cycles, and the other 12 5-cycles. Can you prove this? $\endgroup$ – Krishan Bhalla Apr 29 '15 at 11:28
  • $\begingroup$ The size of a conjugacy class divides the order of the group. There are $24$ five-cycles, and $A_5$ has order $60$ so the class must split. See also math.stackexchange.com/questions/1210503/about-conjugacy-in-a-n/… for some more details $\endgroup$ – Mark Bennet Apr 29 '15 at 11:33
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The conjugacy classes of $A_5$ are the orbits of the action of $A_5$ in $A_5$ given by the conjugacy action. You know that in $S_5$ everyone of those elements are in an unique conjugacy class, and they represents all the classes. You know from the class equation that (in $S_5$)$$|\mathrm{orb}(x)_{S_5}|=\frac{|S_5|}{|\mathrm{Stab}(x)_{S_5}|}$$

We are trying to study $\mathrm{orb}(x)_{A_5}$.

Since $\mathrm{Stab}(x)_{S_5} < S_5$, we have two possibilities:

1) $\mathrm{Stab}(x)_{S_5} \subseteq A_5$: in this case $|\mathrm{orb}(x)_{A_5}|=\frac{1}{2}|\mathrm{orb}(x)_{S_5}|$, so your class in $S_5$ splits in two new classes in $A_5$.

2) $\mathrm{Stab}(x)_{S_5} \nsubseteq A_5$: since $A_5$ is a subgroup of index 2, and since $$\mathrm{Stab}(x)_{A_5}=A_5 \cap \mathrm{Stab}(x)_{S_5}$$ you get$$[\mathrm{Stab}(x)_{A_5}:\mathrm{Stab}(x)_{S_5}]=2$$ So you have $|\mathrm{orb}(x)_{A_5}|=|\mathrm{orb}(x)_{S_5}|$, and you get the same conjugacy class.

Moral: you just have to know if $\exists \tau \in \mathrm{Stab}(x)_{S_5} | \tau \notin A_5$ i.e.: if we are interested in studying if the conjugacy class of $(123)$ splits when you go in $A_5$, bacause $(45) \in \mathrm{Stab}((123))_{S_5} \smallsetminus A_5 $ you know that it DOES NOT SPLIT.

Was I clear?

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  • $\begingroup$ What is this notation? $\exists \tau \in \mathrm{Stab}(x)_{S_5} | \tau \notin A_5$ $\endgroup$ – Permian May 1 '15 at 8:01
  • $\begingroup$ Why do we have two posibilities of $\subseteq A_5$ and $\not\subseteq A_5$? $\endgroup$ – Permian May 1 '15 at 8:03
  • $\begingroup$ I don't understand your question. The notation $\exists \tau... $ in equivalent to $Stab(x) \not\subseteq A_5$. Hope this is helpfull. If it isn't, try to be clearer! $\endgroup$ – Ludox May 1 '15 at 9:33
  • $\begingroup$ We have two possibilities just because they are obviously all the possible cases. It's a bit more difficult to show that, in the second case, the index of Stab(x)_{A_5} in A_5 \cap Stab(x)_{S_5} is 2, but you can try to do this as an exercise. If you need a hint or the solution you just have to ask. $\endgroup$ – Ludox May 1 '15 at 9:37
  • $\begingroup$ Hi, I want to ask the reason of index being $2$...! $\endgroup$ – user152715 Sep 10 '18 at 5:50

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