TITLE: Number of permutations for a cycle-type
QUESTION [11 upvotes]: In the proof that the alternating group $A_5$ is a simple group, we first write cycle-types of the conjugacy classes of $S_5$ which correspond to cycle-types $$1, 2^1, 2^2, 3^1, 2^13^1, 4^1, 5^1$$ and clearly the permutations of cycle-types $1, 2^2, 3^1, 5^1$ are even.
But then my lecture notes state that there is $1$ permutation of cycle type $1$, $15$ of type $2^2$, $20$ of type $3^1$ and $24$ of type $5^1$ making $60$ in total. My question is how did they find the number of permutations for each cycle-type? 
A follow up question is that my lecture notes then state that: 'The problem is that these classes are in $S_n$, and two permutations could conceivably be conjugate in $S_n$ but not in $A_n$.'
Is this because it could be possible that the element $f \in G$ which 'links' two elements in the same conjugacy classes could be in $S_n$ but not in $A_n$?
Thanks!

REPLY [2 votes]: For your first question, here's a start to counting how many cycles of a given type.  Let me show you how to count cycles of type $3^1$ and $2^2$ in $S_4$, and hopefully you can generalize the calculations to your case.  To compute the number of cycles of type $3^1$, first count the number of ways you can pick $3$ numbers to be in the $3$-cycle and $1$ number for the remaining $1$-cycle.  There are ${4 \choose 3} = 4$ ways to do this.  Once you've chosen which numbers to go in your cycles, how many ways can you order the numbers?  Well, in the $1$-cycle, there's clearly only one way.  But in the $3$-cycle, there are $2$ ways.  Why?  Because you can choose any of the $3$ numbers to go in the first position of the cycle, then you have $2$ choices for the number in the second position, and then the third number is set.  That seems like $6$ choices, but you have to remember that you can write the same cycle in $3$ different ways, depending on which number you choose to write first.  So there are really only $3! / 3 = 2$ $3$-cycles that use any particular choice of $3$ numbers.  So to recap, there are $4$ ways to distribute the numbers $1,2,3,4$ into the cycles and for each such distribution, there are $2$ different permutations we can make.  In total, there are $4 \cdot 2 = 8$ cycles of type $3^1$.  If you trace through the argument carefully, you can see that it tells you how to write the cycles down: $(123), (132), (124), (142), \dots$.
For type $2^2$, we first figure how to partition the numbers $1,2,3,4$ into $2$ groups of $2$.  There are ${4 \choose 2} = 6$ ways to do this if we keep track of which order they are in, but we have to divide by $2$ if we don't care about the order, which we don't in the case of cycle types.  The reason we don't care order is that, for example, $(14)(23)$ and $(23)(14)$ represent the same permutation.  So there are only $6 / 2 = 3$ ways to partition the numbers.  And once we do that, there's only one way to produce a $2$-cycle now that we have specified the numbers that go in.  So there are $3$ cycles of type $2^2$: $(12)(34), (13)(24), (14)(23)$.
As for your second question, you are correct.  In general, if you have a group $G$ and a subgroup $H$, and if you consider two elements $h_1, h_2 \in H$, it is possible that they are conjugate in $G$, say $h_2 = g h_1 g^{-1}$ for some $g \in G$, but that there is no such $g \in H$, meaning that they are no longer conjugate in $H$.  Kannapan's link tells you when this occurs in general for the pair $(S_n,A_n)$.