(adsbygoogle = window.adsbygoogle || []).push({}); 1. Let G be a Group, and let H be a subgroup of G. Define the normalizer of H in G to be the set N_{G}(H)= the set of g in G such that gHg^{-1}=H.

a) Prove N_{g}(H) is a subgroup of G

b) In each of the part (i) to (ii) show that the specified group G and subgroup H of G, C_{G}(H)=H, and N_{G}(H)=G

(i) G = D_{4}and H = {1, s, r^{2}, sr^{2}}

(ii) G = D_{5}and H = {1, r, r^{2}, r^{3}, r^{4}}

2. Relevant equations

Notice that if g is an element of C_{G}(H), then ghg^{-1}= h for all h elements of H so, C_{G}(H) is a sub group of N_{G}(H).

3. The attempt at a solution

a) Using the 1 step subgroup test.

If a and b are elements in N_{G}(H) then show ab^{-1}is an element

Let a and b be elements in N_{G}(H), further let a = g= b meaning gHg^{-1}=H=gHg^{-1}. So (gHg^{-1})(gHg^{-1})^{-1}=(gHg^{-1})(g^{-1}H^{-1}g), which by associativity and definition of inverses and closed under inverses, = e which is an element in N_{G}(H), therefor ab^{-1}is an element of N_{G}(H) for every a,b elements of N_{G}(H) and by the one step subgroup test, N_{G}(H) is a subgroup of G.

b) I'm not sure where to begin, or if part a is even right...

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# Homework Help: Groups, Normalizer, Abstract Algebra, Dihedral Groups help?

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