Problem from Artin's Algebra

Question

I want to prove that let S be the set on which group G operates. Let H ={ g∈G | g.s=s for all s∈S} prove that H is normal subgroup of G. This group action gives an homomorphism whose kernal is H. Then it follows directly from statement that H is normal.

How to prove that if the subgroup is the kernel of a homomorphism having the group as its domain then it is normal?

Any help clearing up the statement would be greatly appreciated. Thanks! Is there any other approch ?

Answer 1

Let $\phi$ be a homomorphism. Then for any $k\in\ker\phi$ and $g\in G$, then $$\phi(gkg^{-1})=\phi(g)\phi(k)\phi(g^{-1})= \phi(g)\cdot e\cdot \phi(g)^{-1} =e$$ Therefore $gkg^{-1}\in\ker\phi$, i.e. $\ker\phi$ is normal.

Answer 2

Directly for the situation that you describe:

Note that for $g\in H$ and $x\in G$: $$(xgx^{-1}).s=(xg).(x^{-1}\cdot s)=x.(g.(x^{-1}.s))=x.(x^{-1}.s)=(xx^{-1}).s=e.s=s$$ so that we can conclude that $xgx^{-1}\in H$.