# About Orbit-Stabilizer Theorem, left & right coset, Conjugation

Context: I was playing with Class equation and Orbit-Stabilizer theorem, and I was thinking that If there exist bijection between coset of stabilizer and orbit, conjugating specific element $$a$$ with any element in a coset by stabilizer of $$a$$ should produce the same element ( one of the orbit ).

Let's define group action. the group action $$Conj:G\times G \to G$$ defined by $$Conj(x,y)=xyx^{-1}$$ When there is a Group G and there is a subgroup Stab(a) for $$a\in G$$ and $${e, x_1, x_2, ..., x_k, ..., x_n }\in Stab(a)$$ and there's one of cosets $$\{y_0, y_1, y_2, ..., y_k, ..., y_n\}$$

As I thought according to the existence of bijection between stabilizer cosets and orbit, the following should be true $$Conj(y_k,a)=y_0ay_0^{-1}$$

If $$y_k=y_0x_ke$$ (which means it is left coset) then we can say $$Conj(y_k, a)=y_kay_k^{-1}$$ $$=y_0x_keaex_k^{-1}y_k^{-1}$$ since $$x_keaex_k^{-1}$$ is just conjugation by stabilizer, $$=y_0ay_0^{-1}$$

However in case of defining $$y_k=ex_ky_o$$ (which refers it is right coset)$$Conj(y_k, a) =y_kay_ky^{-1}$$ $$=ex_ky_oay_0^{-1}x_k^{-1}e$$ At this point, It seems it's impossible to sho that $$Conj(y_k, a)=y_0ay_0^{-1}$$

Despite the fact that I can't show $$Conj(y_k, a)=y_0ay_0^{-1}$$ for right coset, Bijection between Orbit and Cosets of stabilizer still holds true since we did check for left coset.

But, since right and left multiplication is just matter of notation, i think we should still be able to check without any contradiction with right coset case.

What is the problem with this checking process? Should I change the conjugation according to defining coset left or rihgt?

(I know that this argument is not about proving Orbit-Stabilizer theorem, But i wanted to check it without any contradiction from theorem) I really wanted to make my question clear as possible but I just couldn't...

• Please use $\operatorname{Conj}(x,y)$ for $\operatorname{Conj}(x,y)$ in future. – Shaun Sep 11 '19 at 19:19
• @Shaun thanks I wil. – WienAudience Sep 12 '19 at 0:26

As your action is a left action, the bijection $$g\cdot a\mapsto g{\rm Stab}_G(a)$$ only necessarily works on left cosets.
If we take for example $$G=S_4$$, $$a=(1,2)$$, $$g=(1,2,3)$$, $$h=(3,4)\in{\rm Stab}_G(a)$$ then $$g\cdot a=gag^{-1}=(2,3)$$ and $$gh\cdot a= gha(gh)^{-1}=(2,3)$$, but $$hg\cdot a=hga(hg)^{-1}=(2,4)$$.
In fact the bijection works for right cosets if and only if $${\rm Stab_G(a)}\trianglelefteq G$$ as for $$h\in {\rm Stab}_G(a)$$ we have $$gh\cdot a=hg\cdot a$$ if and only if $$g\cdot a=hg\cdot a$$ - that is $$h\in{\rm Stab}_G(ga)=g{\rm Stab}_G(a)g^{-1}$$
• Yes, it is common to denote the right action $a^g=g^{-1}ag$ – Robert Chamberlain Sep 12 '19 at 6:01
• In fact (I should have mentioned in my answer) the map $g\cdot a\mapsto a^g=g^{-1}\cdot a$ is a bijection from left actions to right actions which takes $g{\rm Stab}_G(a)$ to ${\rm Stab}_G(a)g^{-1}$ – Robert Chamberlain Sep 12 '19 at 6:29