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Prove that the order of a conjugacy class of $A_5$ is $1,12,15$ or $20$.

There are 4 different cycle type in $A_5$. There are 20 permutations of cycle type $(123)$, 24 permutations of cycle type $(12345)$ and 15 permutations of cycle type $(12)(34)$.

As this is an exercise of the Sylow chapter, I think I need to use it somewhere, but I don't see how. I would say that the centralizers are calculated like:

$C(123)=3$, as I think only $\langle (123) \rangle$ commutes.
$C(12)(34)= 4$, as I think only $\langle \{(12),(34) \} \rangle$ commutes
$C(12345)=5$, as I think only $\langle (12345) \rangle$ commutes

But I don't know how to prove these statements more rigoursly. Is there some kind of theorem to determine the centralizer ?

We know that all elements in a conjugacy class must have the same cycle type. It seems like that the converse is not true for $A_5$. $60/C(12345)=60/5=12$ while there are 24 permutations of this type. For the other cycle types, the conjugacy class doesn't split up. It follows that the order of the conjugacy classes are $60/5=12,\;60/4=15,\;60/3=20$ and $60/60=1$.

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up vote 5 down vote accepted

Suppose that the permutation $g$ commutes with $(1,2,3)$. By the rule for conjugating permutations, we have $(1,2,3) = g(1,2,3)g^{-1} = (g(1),g(2),g(3))$.

Now an $n$-cycle can be expressed as a cycle in $n$ different ways. In the case $n=3$, we have $(1,2,3)=(2,3,1)=(3,1,2)$. So there are three possibilities for $g(1),g(2),g(3): 1,2,3;$ $2,3,1;$ or $3,1,2$. Note that we could also have $g(4),g(5) = 4,5$ or $5,4$, but the only $g$ with $g(4)=5$, $g(5)=4$ are odd permutations, so there are just three possible $g$ in $A_5$: identity, $(1,2,3)$ and $(3,1,2)$.

You can calculate the centralizers of $(1,2)(3,4)$ and $(1,2,3,4,5)$ using similar arguments.

Yes, you are right, there are two classes of 5-cycles in $A_5$. $(1,2,3,4,5)$ and $(1,2,3,5,4)$ are conjugate in $S_5$ but not in $A_5$.

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