Find how many $\tau \in S_{12}$ satisfy $\sigma \tau = \tau \sigma$ for $\sigma = (1\;9\;7\;10\;12\;2\;5)(4\;11)(3\;6\;8) \in S_{12}$

Question

So I've started working on this and I'm not sure if it is correct and also the second question is to actually find those permutations and that really got me stuck.

Let us observe the action of $S_{12} \mapsto S_{12}$ by conjugation.

The objects in $S_{12}$ that interchange with $\sigma$ are exactly the objects in ${\rm Stab}_\sigma$ because

$$\begin{align} \tau \in G_\sigma &\iff \tau \sigma \tau^{-1} = \tau \\ &\iff \tau \sigma = \sigma \tau. \end{align}$$

From the orbit-stabilizer theorem we have $\frac {|S_{12}|}{|G_\sigma|} = |O_\sigma|$.

$O_\sigma$ is exactly all of the objects that are conjugated to $\sigma$ and those are exactly all of the permutations that have the same cycle-formation (as seen in class).

From combinatorics we have:

$$\begin{align} |O_\sigma| &= {{12} \choose {7}}\times 6!\times{{5} \choose {3}}\times 2!\times 1! \\ &= 11,404,800 \end{align}$$

and $|S_{12}| = 12! = 479,001,600$.

So we have $|G_\sigma| = \frac {479,001,600}{ 11,404,800} = 42$.

So first of all I'm not sure this is correct, and secondly I have no idea how to actually find them.

Answer 1

Remark that the first thing to consider when looking for the centralizer(i.e the subgroup consisiting of all permutations commuting with $\sigma$ , which is denoted by $Z(\sigma)$) of an element is to determine the cyclic group it generates, since all powers $\sigma^{k}$ commute with $\sigma$.Now, we recall the fact that if two elements $a,b$ of a group $G$ have finite order and they commute and $Gcd(\vert a\vert ,\vert b\vert)=1$, then $\vert ab \vert =\vert a\vert \times \vert b\vert$. applying this to the disjoint cycles $ (1\;9\;7\;10\;12\;2\;5)$,$(4\;11)$,$(3\;6\;8)$, we deduce that $ \vert(1\;9\;7\;10\;12\;2\;5)(4\;11)(3\;6\;8)\vert=42 $ So the centralizer of $\sigma$ coincide with cyclic group it generates. $$Z(\sigma)=\lbrace ((1\;9\;7\;10\;12\;2\;5)(4\;11)(3\;6\;8))^{k} \backslash k\in \lbrace 0,\cdots 41 \rbrace \rbrace$$ For example,for $k=6$, we find that $Z(\sigma)$ contains $(1\;9\;7\;10\;12\;2\;5)$ and for $k=7$ , it contains $(4\;11)(3\;6\;8)$

Answer 2

If there are $n_r$ cycles of length $r$, we have the formula for counting the number of conjugacy classes

$$ |cl_{S_n}(\sigma)|=\frac{n!}{\displaystyle\prod_{r}r^{n_r}n_r!} $$

Next, use orbit-stablizer theorem, we get the number for the centralizer

$$|C_{S_n}(\sigma)|=\frac{|S_n|}{|cl_{S_n}(\sigma)|}=\prod_{r}r^{n_r}n_r!=(7^1\cdot1!)\cdot(3^1\cdot1!)\cdot(2^1\cdot1!)=42$$