2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I assume that $\;G\curvearrowright X\;$ means "the group $\;G\;$ acts on the set $\;X\;$" ? $\endgroup$ – Timbuc Apr 5 '15 at 15:39
  • $\begingroup$ yes, this is a group action $\endgroup$ – All About Groups Apr 5 '15 at 16:12
  • 1
    $\begingroup$ @Timbuc: that is the "action" version of a sentence like "Let $A \to B$ be a function". $\endgroup$ – Lee Mosher Apr 5 '15 at 18:51
1
$\begingroup$

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

$\endgroup$
-1
$\begingroup$

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):

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.