Question
calculate $\left|C_{S_{12}}(\sigma)\right|$ whereas $\sigma=\left(1...6\right)\left(7...12\right)$
My try
we want to calc the size of the set $\{t\sigma=\sigma t\}$
consider instead $\sigma\mapsto t\sigma t^{-1}$
We know that any permutation that two permutation are replaceable (idk the exact term that means that $a=bab^{-1}$) iff they have the same cycle structure.
therefore we need to find the amount of permutation in $\sigma_{12}$with 2 6-order cycles. this is $\left(\begin{array}{c} 12\\ 6 \end{array}\right)$ .
Now according to orbit- stabilizer that means that
$|S_{12}|=G_{\sigma}$$\left(\begin{array}{c} 12\\ 6 \end{array}\right)$
so $\frac{12!}{\left(\begin{array}{c} 12\\ 6 \end{array}\right)}=G_{\sigma}$
And this is the size I look for?
This is how you can calculate the centralizer of this and any other permutation. Let $\sigma=(1,2,\ldots,6)(7,8,\ldots,12)$. Let $$ \tau= \begin{pmatrix} 1 & 2 & \ldots & 12\\ i_1 & i_2 & \ldots & i_{12} \end{pmatrix}. $$ Then we have (of permutations we multiply from right to left) $$ \tau\sigma\tau^{-1}= \begin{pmatrix} 1 & 2 & \ldots & 12\\ i_1 & i_2 & \ldots & i_{12} \end{pmatrix} (1,2,3,4,5,6)(7,8,9,10,11,12) \begin{pmatrix} i_1 & i_2 & \ldots & i_{12}\\ 1 & 2 & \ldots & 12 \end{pmatrix} $$ $$ =(i_1, i_2, \ldots, i_{6})(i_7, i_8, \ldots, i_{12}). $$ Now $\tau\in C(\sigma)$ if and only if $\tau\sigma\tau^{-1}=\sigma$. Hence we have the equality $$ (1,2,\ldots,6)(7,8,\ldots,12)=(i_1, i_2, \ldots, i_{6})(i_7, i_8, \ldots, i_{12}). $$ If $i_1$ and $i_7$ are given, the remaining symbols $i_k$ are uniquely defined. There are exactly $12$ possibilities for $i_1$. After $i_1$ is chosen for $i_7$ there are exactly 6 possibilities.
So $|C(\sigma)|=12\cdot6$.