Faithful Group Actions and Normal Subgroups

Let $G\curvearrowright X$. Show that $K=\{g\in G:g\cdot x=x,\text{ for all }x\in X \}\trianglelefteq G$. If $\phi\colon G\to Sym(X)$ is the homomorphism given by the action, show that $K=\ker(\phi)$.

Let $G\curvearrowright X$ and let $K$ be as above. Show that $\frac{G}{K}$ acts faithfully on $X$.

Attempt at solution: We want to show that $K = ker(\phi)$ The $ker(\phi)$ has already been show to be a normal subgroup. Also, by definition above we can see that $K$ is not faithful as it has more than one element that acts as the identity.

Take any element $g \in G$ and $x \in X$ Then by definition $g \cdot x = x$ but if we choose different elements $g_1 \in G$ and $x_1 \in X$ then we also get $g \cdot x_1 = x_1$ This is where I am stuck with writing the proof.

In part B when we kill the kernel of $G$ this makes only the identity map to the identity, as this group action now becomes faithful.

• I assume that $\;G\curvearrowright X\;$ means "the group $\;G\;$ acts on the set $\;X\;$" ? – Timbuc Apr 5 '15 at 15:39
• yes, this is a group action – All About Groups Apr 5 '15 at 16:12
• @Timbuc: that is the "action" version of a sentence like "Let $A \to B$ be a function". – Lee Mosher Apr 5 '15 at 18:51

First of all, $G$ may act faithfully on $X$. It was not stated that $K$ had more than one element. Next $g\in K\iff gx=x,\forall\,x\in X\iff \phi(g)=\text{id}_X\iff g\in\ker(\phi)$.

Hints:

For any $\;k\in K\,,\,g\in G\;$ :

$$k\in K\implies k(gx)=gx\;,\;\;\forall\,g\in G\,,\,x\in X\implies$$

$$\implies g^{-1}kg(x)=g^{-1}(kg(x))=g^{-1}gx=1\cdot x= x\implies g^{-1}kg\in K\implies K\lhd G$$

As for (b):

$$(gK)x:=g(x)=x\iff g\in K\iff gK=\overline 1\iff\;\text{the induced action's faithful}$$