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this is my first post here.
I have a question regarding a proof in Algebra by Hungerford:

Let $G$ be a group and $H$ a subgroup of $G$. Let $S$ be the set of all cosets of $H$, where $G$ acts on $S$.
Chapter II Collorary 4.9: "$\dots$ the kernel of $G \rightarrow A(S)$ is a normal subgroup of $G\dots$" Question: Why is the kernel normal?

I looked through the chapter but there were no references, so I figure this is probably "obvious" and hence not included. I came up with an attempt at the proof:

Let $K$ be the kernel of the action.
$ex = x \implies e\in K \implies K$ contains identity.
Let $a,b\in K$.
$(ab)x=a(bx) = ax=x \implies ab\in K \implies K$ is closed.
$a\in K \implies ax=x\implies x=a^{-1}x\implies a^{-1}\in K\implies$ all elements in $K$ has an inverse.
Hence $K$ is a group.

Let $g\in G$.
$(gKg^{-1})x=(gK)(g^{-1}x)=g(K(g^{-1}x))=g(g^{-1}x)=x\implies gKg^{-1}\subset K$
Substitute $g$ for $g^{-1}$ and we get $K\subset gKg^{-1}$.
Hence $gKg^{-1}=K$ and $K$ is normal.

Is this approach correct?

Also, suppose we were to consider the isotropy group of an element $x\in G$ instead.
i.e. $G_x=\lbrace g\in G\mid gx=x\rbrace$
$G_x$ will be a subgroup of $G$ using a similar argument, but now it is not guaranteed to be a normal subgroup because in
$G_x(g^{-1}x)=g^{-1}x$ is not necessary true.
Is this argument right?

Thanks for your time!

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I was unable to tag it as (textbooks) and unable to put "Hi everyone!"... Not sure what is the error, my apologies. – Yong Hao Ng Aug 7 '12 at 18:57
MSE deliberately prevents greetings, salutations, and some "thank yous" and whatnot. But hi to you too, and welcome to the forum! – mixedmath Aug 7 '12 at 18:58
Thanks for the clarification. – Yong Hao Ng Aug 7 '12 at 19:18
up vote 3 down vote accepted

Your proof is fine.

Alternatively, notice that an action is really given by a group homomorphism $G\rightarrow \operatorname{Perm}(S)$ into group of permutations of $S$. Then the kernel of the action is nothing but the kernel of this map, and a kernel of a homomorphism is always normal.

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The alternative way is helpful, thanks for sharing. – Yong Hao Ng Aug 7 '12 at 19:01

The reason the kernel is normal is because given any group homomorphism $\varphi :G\to G'$, the kernel of $\varphi$ is a normal subgroup of $G$, and the way to prove this is exactly the way you did.

Also, as you noted, the stabiliser of a point is always a subgroup, but it does not have to be normal.

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Both are equally good, I chose to accept the one with a higher vote as a tie-breaker. Thanks for your answer too! – Yong Hao Ng Aug 7 '12 at 19:28
@YongHaoNg Fair enough! Both answers were essentially posted at the same time anyway. – M Turgeon Aug 7 '12 at 19:30

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