# Calculation $Stab_G$ of the group $G$

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.

• It should say "The group $S_7$ acts on itself by $x \cdot y = xyx^{-1}$. Calculate..."
– D_S
Commented Jan 20, 2019 at 0:35

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 $$7! \cdot \frac{3^2 2|}{7!}=18$$
• So we get $|G|=7!$ because the order of Symmetric group with $n$ elements is $n!$ and also we get: $|G((123)(456))|=|Orb_{((123)(456))}$| but how to calculate it? Commented Jan 20, 2019 at 0:44
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 $$S_n$$.