Thread: Kernel of a group action

This might be a silly question, but I'm trying to figure out why the kernel of a group action is defined the way it is. If G is a group acting on a set A, the kernel of the action is defined as {g in G | ga=a for all a in A}, which is also equivalent to the intersection of all the stabilizers.

In general, the kernel of a map is the set of things that are mapped to the identity. Is there some way in which that definition matches up with the above definition? I'm sure it must, but I can't quite see how. I can see that if ga=a for all a then that is sort of the identity map, so maybe that's why, but it isn't the map that maps everything to the identity. Is there some other way of looking at a group action that makes this be the standard definition of the kernel?

Thanks.

Originally Posted by Hypatia

This might be a silly question, but I'm trying to figure out why the kernel of a group action is defined the way it is. If G is a group acting on a set A, the kernel of the action is defined as {g in G | ga=a for all a in A}, which is also equivalent to the intersection of all the stabilizers.

In general, the kernel of a map is the set of things that are mapped to the identity. Is there some way in which that definition matches up with the above definition? I'm sure it must, but I can't quite see how. I can see that if ga=a for all a then that is sort of the identity map, so maybe that's why, but it isn't the map that maps everything to the identity. Is there some other way of looking at a group action that makes this be the standard definition of the kernel?

Thanks.

"kernel" here means the kernel of the homomorphism defined by for all and

Do you mean $\phi$ from $GXA$ to $Sym(A)$ (that is the same as $S_{A}$ right?). Because we need two inputs, g and a, not just g?

(Darn, how did you make the latex show up properly?)

Originally Posted by Hypatia

Do you mean $\phi$ from $GXA$ to $Sym(A)$ (that is the same as $S_{A}$ right?). Because we need two inputs, g and a, not just g?

(Darn, how did you make the latex show up properly?)

don't confuse with the "action function" which is from to as i said, is from to or the group of permutations over .

so for any is a bijection from onto proving that is a group homomorphism and is straightforward.

Thanks for your help. It makes more sense now that its being mapped to the identity permutation, so the choice of the word kernel makes sense.

The only thing I'm still confused about is the notation you're using with phi(g)(a). Is that phi applied to g and then to a? And I get that if a goes to a for all a then the permutation going on there is the identity permutation, but how does the notation ga fit in as a permutation?

Originally Posted by Hypatia

Thanks for your help. It makes more sense now that its being mapped to the identity permutation, so the choice of the word kernel makes sense.

The only thing I'm still confused about is the notation you're using with phi(g)(a). Is that phi applied to g and then to a? And I get that if a goes to a for all a then the permutation going on there is the identity permutation, but how does the notation ga fit in as a permutation?

for a fixed and any is the image of under the map of course, the first thing you need to show is that is well-defined, i.e. for any the map is basically

a bijection from onto this is quite easy to see: so is injective. also for let then:

which proves that is surjective and hence next show that is a homomorphism: for all and

therefore finally