How can I prove that $C_{S_4}((12)(34))$ is a subgroup of order 8?

Question

I would like to know if there is a smart method (i.e. avoid to make all by hand) to understand the order of $C_{S_4}((12)(34))$. The only thing I thought is that $|S_4:C_{S_4}((12)(34))|=|\mathcal{O}((12)(34))|$ and then one can try to reason about the order of the orbit. But this didn't helped me! Thank you

Answer 1

The conjugacy class of $(12)(34)$ consists of all products of disjoint transpositions, since two permutations are conjugate if and only if they have the same cycle type. So, only $(13)(24)$ and $(14)(32)$ are conjugate to $(12)(34)$. As you mention, the size of the conjugacy class equals the index of the stabilizer of $(12)(34)$. Hence, the stabilizer has index $3$, and therefore order $8$.

Answer 2

There is an easy way by using combinatorics. We need to count elements in the form $(a,b)(c,d)$;

$ {1}\over{2}$ $ {4}\choose{2} $$2\choose 2$ $=3$ we divided by $2$ since $(a,b)(c,d)=(c,d)(a,b)$

So answer is $24\over 3$=$8$ by orbit stabilizer theorem.

You can even do it in larger groups,if you want to find same thing in $S_6$ then;

$1\over 2$$6\choose2 $$4\choose2$ $=45$ then by orbit stabilizer theorem ${6!}\over{45}$ $=16$

Note that I used the fact that conjuge class of an element in $S_n$ is set of all elements having same permutation cylic type.