# Kernel of a group action

• Oct 19th 2011, 02:53 AM
Bernhard
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 $\in$ A.
Show that the following sets are subgroups of G:
(a) the kernel of the action
(b) {g $\in$ 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 $\in$ G| ga = a for all a $\in$ A}

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

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

Then $k_1$a = a and $k_2$a = a

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

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

Now ( $k_1$ ${k_2}^{-1}$)a = $k_1$( ${k_2}^{-1}$a)

But ??? where to from here

Can anyone help? Be grateful for some assistance!

Peter
• Oct 19th 2011, 02:58 AM
slevvio
Re: Kernel of a group action
Hello,

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

is that cool?
• Oct 19th 2011, 03:01 AM
Bernhard
Re: Kernel of a group action
Thanks for the excellent response - so quickly!!

Great mate!

Peter
• Oct 19th 2011, 03:02 AM
slevvio
Re: Kernel of a group action
no problem!