# Groups, Normalizer, Abstract Algebra, Dihedral Groups help?

1. Oct 8, 2008

### nobody56

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}

2. Relevant 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).

3. 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...

2. Oct 8, 2008

### Dick

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)?

3. Oct 8, 2008

### nobody56

Would c^(-1)=(ab^(-1))^(-1)=a^(-1)b?

4. Oct 8, 2008

### nobody56

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 lets 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)?

5. Oct 8, 2008

### Dick

Mind the ordering. (ab^(-1))^(-1)=ba^(-1). To prove it multiply that by ab^(-1). Do you see how that works?

6. Oct 8, 2008

### Dick

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?

7. Oct 8, 2008

### nobody56

yeah, i forgot the ordering....as for c^(-1)Hc=(c^(-1)Hc)^(-1)=cHc^(-1) by the same ordering property right?

8. Oct 8, 2008

### Dick

Ok, since H^(-1)=H.

9. Oct 8, 2008

### nobody56

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?

10. Oct 8, 2008

### Dick

Yes, now you just have to do some calculations in the dihedral groups to try and figure out what the normalizer and centralizer are.

11. Oct 10, 2008

### nobody56

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

12. Oct 10, 2008

### Dick

I would think you could just state the answers without showing every calculation you did. But that's just my opinion.

13. Oct 10, 2008

### nobody56

i decided to play it safe, and just wrote it all out, thank you for your time and help!