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How do we efficiently calculate the number of elements of a specific cycle-type in groups such as $\operatorname{Alt}(n)$ (Alternating group of $n$ letters) and $\operatorname{Sym}(n)$ (Symmetric groups of $n$ letters)?

I know one way that involves calculating the cardinality of the conjugacy class that contains elements of the same cycle type (elements of the same cycle type are conjugate and so they belong to same conjugacy class). Two elements being conjugate is an equivalence relation, and so elements of each cycle type will be partitioned into separate classes. By looking at the cardinality of each conjugacy class (or combined split classes in the alternating group), we can determine the number of elements of a specific cycle type.

Each conjugacy class has size $[G:C_G(x)]$ where $C_G(x)$ is the set of all elements that commute with $x \in X.$ We have that $|\operatorname{Alt}(4)| = \frac{n!}{2} = \frac{24}{2} = 12,$ and $|C_G((12)(34)) |= 2 \cdot 2 \cdot 2 = 8.$ Thus $|\operatorname{Alt}(4)|/|C_G((12)(34))| = 24/8 = 3,$ which means that $\operatorname{Alt}(4)$ contains $3$ disjoint $2$-cycles. Is there a quicker way to do this?

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For a simple method, use combinatorics. For calculating the number of $3$-cycles in $\operatorname{Alt}(4)$, we can just multiply the number of unique choices in each slot and didide by the order of that particular cycle type. So $\frac{4 \cdot 3 \cdot 2 }{3} = \frac{24}{3}.$ To calculate say the number of $5$-cycles in $\operatorname{Sym}(7)$, we do the following calculation: $\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{5}$

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In $S_n$, a cycle type is just a partition $\lambda \vdash n$, for instance a valid cycle type for $S_6$ could be $(3,2,1)$, and a permutation with this cycle type could be $(123)(45)$. Using the notation $\lambda = (1^{a_1},2^{a_2},\ldots,n^{a_n})$ with $\sum i a_i = n$, we can define $c_\lambda$ to be the cycles with cycle type $\lambda$. Then we see the following:

An arbitrary permutation consists of an ordering of numbers and a placement of brackets. With our cycle type set, the brackets are in place. We can naively count that there are $n!$ ways to order the numbers in the brackets, and then account for double counting. For each $j$-cycle, we have $j$ ways to write it that yield the same cycle; for instance $(123) = (231) = (312)$. Thus we need to divide by each $(j)^{a_j}$ for each $j$. Moreover, if we have more than one $j$-cycle, then we have $a_j!$ ways to order the $j$-cycles; for instance $(12)(34) = (34)(12)$. Putting this together yields $$ |c_{\lambda}| = \frac{n!}{\prod_j (j)^{a_j} (a_j!)} $$

In the case you wrote out, we are looking for $|c_{(0,2)}|$ in $S_4$, so we have $$ |c_{(0,2)}| = \frac{4!}{2^2 \cdot 2!} = 3 $$ as you calculated. This formula appears in many combinatorics textbooks (certainly in Sagan's text on the Symmetric Group, and likely in one of Stanley's texts) and is listed on groupprops.

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