Kernel of a group action

**Hypatia** · Oct 10th 2009, 06:44 PM

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.

**NonCommAlg** · Oct 10th 2009, 06:59 PM

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 $\varphi : G \longrightarrow \text{Sym}(A)$ defined by $\varphi(g)(a)=ga$ for all $g \in G$ and $a \in A.$

**Hypatia** · Oct 10th 2009, 07:05 PM

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?)

**NonCommAlg** · Oct 10th 2009, 07:24 PM

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 $\varphi$ with the "action function" which is from $G \times A$ to $A.$ as i said, $\varphi$ is from $G$ to $\text{Sym}(A)$ or $S_A,$ the group of permutations over $A$ .

so for any $g \in G, \ \varphi(g)$ is a bijection from $A$ onto $A.$ proving that $\varphi$ is a group homomorphism and $\ker \varphi = \{g \in G: \ ga=a, \ \forall a \in A \}$ is straightforward.

**Hypatia** · Oct 10th 2009, 07:27 PM

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?

**NonCommAlg** · Oct 10th 2009, 07:55 PM

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 $g \in G$ and any $a \in A, \ \varphi(g)(a)$ is the image of $a$ under the map $\varphi(g).$ of course, the first thing you need to show is that $\varphi$ is well-defined, i.e. for any $g \in G,$ the map $\varphi(g)$ is basically

a bijection from $A$ onto $A.$ this is quite easy to see: $\varphi(g)(a)=\varphi(g)(b) \Longleftrightarrow ga=gb \Longleftrightarrow a = b.$ so $\varphi(g)$ is injective. also for $b \in A$ let $a=g^{-1}b.$ then: $\varphi(g)(a)=ga=g(g^{-1}b)=(gg^{-1})b=b,$

which proves that $\varphi(g)$ is surjective and hence $\varphi(g) \in S_A.$ next show that $\varphi$ is a homomorphism: $\varphi(g_1g_2)(a)=(g_1g_2)a=g_1(g_2a)=\varphi(g_1 )(g_2a)=\varphi(g_1)\varphi(g_2)(a),$ for all $g_1,g_2 \in G$ and $a \in A.$

therefore $\varphi(g_1g_2)=\varphi(g_1) \varphi(g_2).$ finally $\ker \varphi = \{g \in G: \ \varphi(g) = \text{identity map} \} = \{g \in G: ga = \varphi(g)(a)=a, \ \forall a \in A \}.$

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