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nobody56
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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 NG(H)= the set of g in G such that gHg-1=H.
a) Prove Ng(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, CG(H)=H, and NG(H)=G
(i) G = D4 and H = {1, s, r2, sr2}
(ii) G = D5 and H = {1, r, r2, r3, r4}
Notice that if g is an element of CG(H), then ghg-1 = h for all h elements of H so, CG(H) is a sub group of NG(H).
a) Using the 1 step subgroup test.
If a and b are elements in NG(H) then show ab-1 is an element
Let a and b be elements in NG(H), further let a = g= b meaning gHg-1=H=gHg-1. So (gHg-1)(gHg-1)-1=(gHg-1)(g-1H-1g), which by associativity and definition of inverses and closed under inverses, = e which is an element in NG(H), therefor ab-1 is an element of NG(H) for every a,b elements of NG(H) and by the one step subgroup test, NG(H) is a subgroup of G.
b) I'm not sure where to begin, or if part a is even right...
a) Prove Ng(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, CG(H)=H, and NG(H)=G
(i) G = D4 and H = {1, s, r2, sr2}
(ii) G = D5 and H = {1, r, r2, r3, r4}
Homework Equations
Notice that if g is an element of CG(H), then ghg-1 = h for all h elements of H so, CG(H) is a sub group of NG(H).
The Attempt at a Solution
a) Using the 1 step subgroup test.
If a and b are elements in NG(H) then show ab-1 is an element
Let a and b be elements in NG(H), further let a = g= b meaning gHg-1=H=gHg-1. So (gHg-1)(gHg-1)-1=(gHg-1)(g-1H-1g), which by associativity and definition of inverses and closed under inverses, = e which is an element in NG(H), therefor ab-1 is an element of NG(H) for every a,b elements of NG(H) and by the one step subgroup test, NG(H) is a subgroup of G.
b) I'm not sure where to begin, or if part a is even right...