Find all permutations that commute with $\omega$=(1 9 7 10 12 2 5)(4 11)(3 6 8) in $S_{12}$

Solution 1:

Your reasoning is fine, in my opinion. $\sigma\in S_{12}$ commutes with $\omega$ if and only if $\sigma\omega\sigma^{-1} = \omega$. Since there are $7\cdot 2\cdot 3$ ways to fix a representation of $\omega$ in the form $$ \sigma\omega\sigma^{-1} = (\sigma(1)\ \sigma(9)\ \sigma(7)\ \sigma(10)\ \sigma(12)\ \sigma(2)\ \sigma(5))\ (\sigma(4)\ \sigma(11))\ (\sigma(3) \ \sigma(6)\ \sigma(8)), $$ and each such representation uniquely determines $\sigma$, there are exactly $7\cdot 2\cdot 3$ elements $\sigma\in S_{12}$ which commute with $\omega$.

Here is an alternative solution using the theory of group actions to determine the number of elements of $S_{12}$ which commute with $\omega$:

The question asks for the centralizer $C(\omega)$ of $\omega$ in $S_{12}$. Now look at the group operation of $G$ on itself by conjugation. Then $C(\omega)$ is the stabilizer of $\omega$. The orbit $O$ of $\omega$ is the conjugacy class of $\omega$ in $S_{12}$, which is the set of all elements of the same cycle type. The standard counting method for permutations yields $$\left|O\right| = \frac{12!}{2\cdot 3\cdot 7}.$$

Now by the orbit-stabilizer-theorem, $\left|C(\omega)\right| = \left|S_{12}\right|/\left|O\right| = 2\cdot 3\cdot 7$.