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I tried to derive it and I got 60 = 1+15+20+24. But we know that for every a in G; o(a)/o(G) and 24 does not divide 60 then how it is possible class equation. Your small hint might be helpful to me. Thanks.

Sonu Lamba
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    You may want to explain how you have derived your equation, perhaps there is just a minor mistake there. – Wojowu Jul 01 '18 at 15:58
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    While it is true that all 5-cycles are conjugate in $S_5$ they are not all conjugate in $A_5$. You have just discovered evidence of that happening (+1 for that)! – Jyrki Lahtonen Jul 01 '18 at 15:58
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    One way of thinking about it is the following. Let $\alpha$ be a 5-cycle. You know that it has $24$ conjugates in the group $S_5$. Therefore $C_{S_5}(\alpha)$ must have order five. But, $\alpha$ commutes with itself, and generates a group of order five. Therefore $C_{S_5}(\alpha)=\langle\alpha\rangle$. That centralizer is contained in $A_5$, so we see that $C_{A_5}(\alpha)$ is also of order five. The orbit-stabilizer theorem then tells that $\alpha$ has $60/5=12$ conjugates in $A_5$. In other words, the $S_5$ conjugacy class splits in two. – Jyrki Lahtonen Jul 01 '18 at 16:02
  • @Wojowu can you please tell me that minor mistake. Actually I am not able to attach the pic. And not good at TeX codes. Please help & explain the problem. 24 does not divide 60 then how can it be possible as class equation of A5. Thanks – Sonu Lamba Jul 01 '18 at 16:03
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    The general result is that a conjugacy of $S_n$ either splits in two or remains a class in $A_n$. All depending on whether the centralizer (in $S_n$) consists of only even permutations or also has odd permutations. – Jyrki Lahtonen Jul 01 '18 at 16:04
  • @SonuLamba I'm afraid I cannot tell you what your mistake is if I don't know what your reasoning was. – Wojowu Jul 01 '18 at 16:05
  • Start from here. There are links to several threads where this topic is covered quite generally. – Jyrki Lahtonen Jul 01 '18 at 16:07
  • This looks also relevant. I'm having a hard time picking a dupe target. – Jyrki Lahtonen Jul 01 '18 at 16:17
  • Thanks @Jyrki Lahtonen . Your efforts are helpful. – Sonu Lamba Jul 01 '18 at 16:22

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The class equation would be $60=1+20+12+12+15$, instead of $60=1+20+24+15$. There are $24$ cycles of length $5$ here, all of which are not conjugate to each other. Actually there are two different conjugacy classes each of size $12$, e.g., $(12345)$ and $(13524)$ are not in the same conjugacy class.

Xander Henderson
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Heerack Ganguly
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