Question #300576

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


1
Expert's answer
2022-02-24T05:40:54-0500

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

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

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

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

It also follows from ha=ahha=ah that a1haa1=a1aha1,a^{-1}haa^{-1}=a^{-1}aha^{-1}, and hence a1h=ha1.a^{-1}h=ha^{-1}. We conclude that a1N(a).a^{-1}\in N(a).

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

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