# Calculating the orbit of a group

In the $$S_7$$ group, in action on itself by $$xy=xyx^{-1}$$, I would like to calculate the orbit of $$(123)(456)$$.

I read the definition:

Consider a group $$G$$ acting on a set $$X$$. The orbit of an element $$x$$ in $$X$$ is the set of elements in $$X$$ to which $$x$$ can be moved by the elements of $$G$$. The orbit of $$x$$ is denoted by $$G\cdot x$$: $$G\cdot x = \{g\cdot x | g\in G\}$$

But I can't seem to understand how to calculate technically speaking the orbit of $$(123)(456)$$. My goal is to calculate $$|Stab_{S_7}(123)(456)|$$. In order to achieve it I have to calculate the orbit, but how?

• What action are you studying? Presumably $S_7$ acting on itself, but by conjugation, or one of the regular actions? Jan 20, 2019 at 12:52
• @user3482749 Sorry, I meant on itself, ill edit. Jan 20, 2019 at 12:57
• There is more than one action of $S_7$ on itself. Do you mean the action by conjugation? Jan 20, 2019 at 13:00
• @user3482749 on itself by $xy=xyx^{-1}$ Jan 20, 2019 at 13:01

If you have a permutation written down as its cycle decomposition, it's very easy to describe the action of an element by conjugation - that element is applied to all of the entries in the cycles. If $$y$$ includes the cycle $$(a_1a_2\dots a_k)$$, $$\sigma y\sigma^{-1}$$ includes the cycle $$(\sigma(a_1)\sigma(a_2)\dots\sigma(a_k))$$.
So then, we start with an element that's a composition of two disjoint $$3$$-cycles. Conjugate by something, and we get a composition of two disjoint $$3$$-cycles. Can you see what the orbit will be now?