Answer to Question #300576 in Abstract Algebra for Jp@88

Question

PROVE that N (a) is a subgroup of G Where N (a) is a normalizer of a in G

Answer 1

Let us prove that a normalizer $N (a)=\{g\in G: ga=ag\}$ of $a$ is a subgroup of $G.$

Let $h,k\in N(a).$ Then $ha=ah$ and $ka=ak.$ It follows that

$(hk)a=h(ka)=h(ak)=(ha)k=(ah)k=a(hk),$

and hence $hk\in N(a).$

It also follows from $ha=ah$ that $a^{-1}haa^{-1}=a^{-1}aha^{-1},$ and hence $a^{-1}h=ha^{-1}.$ We conclude that $a^{-1}\in N(a).$

Therefore, $N (a)$ is a subgroup of $G.$