TITLE: Counting cycle structures in $S_n$
QUESTION [10 upvotes]: Is there a fast way to count all the different types of cycle structures in a Symmetric group?
More specifically: "How many elements are there in $S_8$ of cycle structure $4^2$"
Here, $4^2$ means a permutation in $S_8$ that is the product of 2 4-cycles i.e: 
( 1 2 3 4 )( 5 6 7 8 ) would be an element. ( 2 4 6 8)(1 3 5 7) would also be an element. Is there a general formula that would tell me the number of elements in this set? Or do I have to just have to calculate each one? Thanks!

REPLY [6 votes]: You can count the cycle structure in $S_n$ using combinatorial techniques. For your example, in $S_8$ you want a product of two 4-cycles: $(a_1 \, a_2 \, a_3 \, a_4)(a_5 \, a_6 \, a_7 \, a_8)$. 
Start by counting the number of ways to set up the leftmost cycle. You have 8 choices for $a_1$, 7 choices for $a_2$, 6 choices for $a_3$ and 5 choices for $a_4$. But we have over counted. Notice that for any 4-cycle in $S_8$ $(a_1 \, a_2 \, a_3 \, a_4)=(a_2 \, a_3 \, a_4 \, a_1)=(a_3 \, a_4 \, a_1 \, a_2)=(a_4 \, a_1 \, a_2 \, a_3)$. This tells us that we have over counted by 4 for the first 4-cycle so we must divide by 4. This yields $\frac{8\cdot7\cdot6\cdot5}{4}$ possible 4-cycles for the leftmost cycle. 
Next count the number of ways to set up the rightmost cycle. You have already used up 4 out of 8 numbers available so you only have 4 choices for $a_5$, 3 choices for $a_6$, 2 choices for $a_7$ and 1 choice for $a_8$. Again, we have over counted by four so we have $\frac{4\cdot 3\cdot 2 \cdot 1}{4}$ possible 4-cycles for the rightmost cycle. 
Finally, since $(a_1 \, a_2 \, a_3 \, a_4)(a_5 \, a_6 \, a_7 \, a_8)=(a_5 \, a_6 \, a_7 \, a_8)(a_1 \, a_2 \, a_3 \, a_4)$ we need to divide by 2 (since right now we are counting say (1 2 3 4)(5 6 7 8) and (5 6 7 8)(1 2 3 4) separately).
So in $S_8$ we have $\frac{8 \cdot 7\cdot 6 \cdot 5}{4} \cdot \frac{4 \cdot 3 \cdot 2 \cdot 1}{4} \cdot \frac{1}{2}=\frac{8!}{4^2 2!}$. 
(This coincides with the formula you have already been given.)