How to use group actions to prove Lagrange's theorem?

Question

I was browsing proof wiki and saw that is possible to use group actions to prove Lagrange's theorem as an immediate corollary of the orbit stabiliser theorem but I don't quite see how it follows. I have so far defined the group action (on the left coset space) as given on the page (except I used $gH$ for a left coset whereas they just use $H$ for a arbitrary left coset (presumably this doesn't matter then?)).

I have proved using my action $\phi:G\times G/H \rightarrow G/H$ by $\phi:(g_1,g_2H) \mapsto (g_1g_2)H$ I have shown this is a group action.

So from the orbit stabiliser theorem we have $|O_{gH}|=|G|/|\text{Stab}_G(gH)|$ I suspect that $|\text{Stab}_G(gH)|=|H|$ and $|O_{gH}|=[G:\text{Stab}_G(gH)]$ but I'm not fully sure of this and even then how can I finish this proof?

Answer 1

You just need the orbit of one left coset.

The stabilizer of $H$ is $\{x\in G: xH=H\}$. Now $xH=H$ if and only if $x\in H$.

Answer 2

For the neutral element $H=e_{G/H}$ of the semi-group $G/H$, it is easy to calculate $Stab_{G}(H)=\{g\in G| gH=H\}=H$ as mentioned. Now from Orbit-Stabilisator theorem one has:

$$G/H=G/Stab_{G}(H)\cong_{Set}O_{H}=\{gH|g\in G\}$$

which means in particular that: $[G:H]:=|G/H|=|O_{H}|$

the whole goal of Lagrange's theorem is to show [G:H]=|G|/|H|, I don't know which fact (without using Lagrange's theorem) you used to have

$$|G/Stab_{G}(H)|=|G|/|Stab_{G}(H)|$$

unless you want to stay on your circular argument, the usual proof goes this way we act $G$ on $H$ by translation:

$$\phi: G\longrightarrow Aut_{Set}(H)$$ where $$\phi(g):H\longrightarrow H$$ with $\phi(g)(h)=g\cdot_{G} h$.

Clearly the inverse is given by $\phi(g^{-1})$. In this action, the orbit $O_g=gH$ have same cardinality of H. Now since in every group action the set of orbit partition the set on which we act (we show this by showing it is an equivalence relation) this means in our case that it exists a set $\overline{G}\subset G$ such that $G=\bigsqcup_{g\in\overline{G}}gH$, in fact $|\overline{G}|=|O_H|$. Hence

$$|G|=|\bigsqcup_{g\in\overline{G}}gH|=\sum_{g\in\overline{G}}|gH|=\sum_{g\in\overline{G}}|H|=|H||O_H|=|H||G/H|$$

as desired.