Calculation $Stab_G$ of the group $G$

Question

So, I was trying to understand the "Group action" theory. I read the definition of $Stab_G$ and I was trying to solve some basic questions.

I came across with the following question:

Let $S_7$ be a group that on itself by $x\cdot y = xyx^{-1}$. Calculate $|Stab_{S_7}((1 \ 2 \ 3)(4 \ 5 \ 6))|$.

Firstly, I don't understand what does "on itself by $x\cdot y = xyx^{-1}$" means. Secondly I would like to see how to calculate it formally so I can calculate other sections of the question.

Answer 1

A group action on a set $X$ is defined as $\varphi : G\ \mathrm{x}\ X \rightarrow X$ such that $g \cdot x=\varphi (g,x)$ In this case $X=G$ and $\varphi (x,y)=xyx^{-1}$
Now $$|Stab_{S_7}((1 \ 2 \ 3)(4 \ 5 \ 6))|=\frac{|G|}{|G((123)(456))|}$$ where $G((123)(456))$ is the orbit of $(123)(456)$ under the action of $G$
The orbit of an element in $S_7$ is determined by its cycle type: in this case is $(3,3,1)$, so $|G((123)(456))|$ is the number of $(3,3,1)$ cycles in $S_7$. The order of its conjugacy class is given by the formula in the link: https://groupprops.subwiki.org/wiki/Conjugacy_class_size_formula_in_symmetric_group
In our case $$|G((123)(456))|=\frac{7!}{3^2 2!}=280$$ and the order of the stabilizer is $7! \cdot \frac{3^2 2|}{7!}=18$

Answer 2

One way to proceed is to try to improve our understanding of the action. For instance, if we write $y = (1\ 2\ 3)(4\ 5\ 6)$, what is $(x\cdot y)(x(1))$, the image of $x(1)$ under $x\cdot y$? how about $x(2)$? In general, you should likely have seen some theorem about what conjugation does to a cycle in $S_n$.