Groups Theory. Prove the intersection of stabilizer is exactly the kernel of the action.

Question

Prove that $ \cap_{x\in X}G_x $ is exactly the kernel of the action

The first question I have is that what is the action.

I assumed that the action is $\phi:G \longrightarrow X $ where $\phi(g)=gx$

Now $ Ker\phi=${$g\in G | \phi(g)=0$}

But I don't know what is $ \cap_{x\in X}G_x $

I think that I don't understand anything of these things but I have googled and read my notes and I cannot do anything...

Answer 1

Any action of a group on a set determines a homomorphisn from that group to the symmetries group on that set (and the other way around), and it is this homomorphism's kernel that we call "the action's kernel".

The kernel is thus the set of all elements in $\;G\;$ which act as the identity permutation on $\;X\;$:

$$\ker\phi:=\left\{\,g\in G\;:\;\forall\,x\in X\;,\;\;gx=x\,\right\}=\left\{\;g\in G\;;\;g\in G_x\;,\;\;\forall\,x\in X\;\right\}=$$

$$=\bigcap_{x\in X} G_x$$