How many disjoint products of two $2$-cycles are there in $S_5$?

Question

How many disjoint products of two $2$-cycles are there in $S_5$ ?

In general I'm having trouble in determining number of disjoint products of cylces. Please help. Thanks in advance.

Answer 1

There are different ways to arrive at the correct expression. Here is my try:

• first you realize that only 4 out of the 5 symbols appear in the cylce $(\cdot, \cdot) (\cdot,\cdot)$. There are ${5\choose 4}=5$ to choose the 4 symbols.

• The four symbols chosen in the first step can by distributed in $4!$ ways on the four slots in the cycle. However, $(1,2) =(2,1)$ so we overcount by a factor of 2 in each of the cylces. Additionally, we overcount by another factor of 2 due to the fact that $(1,2)(3,4) = (3,4)(1,2)$, that is exchanging the two cylces leads to the same element. So in total there are $4!/2^3$ ways to distribute the 4 elements in the cycle structure and achieving distinct elements of $S_5$.

• So in total we have $$ {5\choose 4} \frac{4!}{2^3} = 15 $$ elements.

Answer 2

We have a permutation of four elements, as the last element is fixed: ($\_$ $\_)$ $(\_$ $\_)$. We permute these elements in $4!$ ways. Now the first two elements commute and the second two elements commute because they are two-cycles. So we divide out by $4$. This gives us $3!$ permutations. However, disjoint cycles commute, so we divide out again by $2$, leaving us with $3$ permutations.

Since we have five elements, we pick four to permute in $\binom{5}{4} = 5$ ways. This is independent of arranging the elements, so we multiply $5 \cdot 3 = 15$ by rule of product.