Proving normalizer is a subgroup and computing centralizers for dihedral groups

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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}

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...
 

Discussion

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The proof is not very good. You can't assume a=b! You take a and b in NG(H). So aHa^(-1)=H and bHb^(-1)=H. You want to show c=ab^(-1) is in NG(H). Which means cHc^(-1)=H. First off, what is c^(-1)?
 
Would c^(-1)=(ab^(-1))^(-1)=a^(-1)b?
 
could i say let a, b be elements of NG(H), and let a = aHa^-1 and b = bHb^(-1), and since a and b are elements in NG(H) aHa^(-1)=bHb^(-1), then use the division algorithm for right cancellation to say aHa^(-1)b=bHb^(-1)b which goes to aHc^(-1)=bHe and then similarly by the division algorithm for left cancellation, a^(-1)aHc^(-1)=a^(-1)bHe, which simplifies to eHc^(-1)=c^(-1)He...would that let's us say c^(-1) is and element in NG(H), getting us to c^(-1)H(c^(-1))=c^(-1)Hc...but can i claim commutativity and say cHc^(-1)?
 
nobody56 said:
Would c^(-1)=(ab^(-1))^(-1)=a^(-1)b?

Mind the ordering. (ab^(-1))^(-1)=ba^(-1). To prove it multiply that by ab^(-1). Do you see how that works?
 
nobody56 said:
could i say let a, b be elements of NG(H), and let a = aHa^-1 and b = bHb^(-1), and since a and b are elements in NG(H) aHa^(-1)=bHb^(-1), then use the division algorithm for right cancellation to say aHa^(-1)b=bHb^(-1)b which goes to aHc^(-1)=bHe and then similarly by the division algorithm for left cancellation, a^(-1)aHc^(-1)=a^(-1)bHe, which simplifies to eHc^(-1)=c^(-1)He...would that let's us say c^(-1) is and element in NG(H), getting us to c^(-1)H(c^(-1))=c^(-1)Hc...but can i claim commutativity and say cHc^(-1)?

You cannot say a=aHa^-1 for a start. H=aHa^-1. Not a. The rest of it is sort of ok, If you can get to c^(-1)Hc=H that's fine, you don't need any commutativity to turn that into cHc^(-1)=H. Do you see why?
 
yeah, i forgot the ordering...as for c^(-1)Hc=(c^(-1)Hc)^(-1)=cHc^(-1) by the same ordering property right?
 
Ok, since H^(-1)=H.
 
right, so for b could I take the definition of a Centralizer CG(H)={g element of G, such that ga=ag} and just plug in the elements to show CG(H)=H? and do the same for the normalizer?
 
Yes, now you just have to do some calculations in the dihedral groups to try and figure out what the normalizer and centralizer are.
 
So then would i just be able to say that since CD4(r2)={1, r2, s, sr2},and CD4(s)={1, r2, s, sr2}, and CD4(sr2)={1, r2, s, sr2}, then CD4(1, s, r2, sr2)={1, r2, s, sr2}...or should i try and show each part, as in 1*r=r*1, which seems kinda tidious
 
I would think you could just state the answers without showing every calculation you did. But that's just my opinion.
 
i decided to play it safe, and just wrote it all out, thank you for your time and help!