# Thread: Kernel of a group action

1. ## Kernel of a group action

Dummit and Foote Section 1.7 Group Actions - Exercise 4 is as follows:

Let G be a group acting on a set A and fix some a $\displaystyle \in$ A.
Show that the following sets are subgroups of G:
(a) the kernel of the action
(b) {g $\displaystyle \in$ G | ga = a} {stabilizer of a in G}

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I am working on part (a) and have not got far!

We know that the kernel of the action of G on A is as follows:

Kernel, K = {g $\displaystyle \in$ G| ga = a for all a $\displaystyle \in$ A}

We need to show x$\displaystyle y^{-1}$ $\displaystyle \in$ K for all x, y $\displaystyle \in$ K

So assume $\displaystyle k_1$, $\displaystyle k_2$ $\displaystyle \in$ K

Then $\displaystyle k_1$a = a and $\displaystyle k_2$a = a

Now need to show $\displaystyle k_1$$\displaystyle {k_2}^{-1}$$\displaystyle \in$ K

That is ($\displaystyle k_1$$\displaystyle {k_2}^{-1})a = a Now (\displaystyle k_1$$\displaystyle {k_2}^{-1}$)a = $\displaystyle k_1$($\displaystyle {k_2}^{-1}$a)

But ??? where to from here

Can anyone help? Be grateful for some assistance!

Peter

2. ## Re: Kernel of a group action

Hello,

$\displaystyle k_2 a = a \implies k_2^{-1} k_2 a = k_2 ^{-1} a \implies k_2^{-1} a = a$ ...

is that cool?

3. ## Re: Kernel of a group action

Thanks for the excellent response - so quickly!!

Great mate!

Peter

no problem!