Groups, Normalizer, Abstract Algebra, Dihedral Groups help?

[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 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 = 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 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).

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^-1H^-1g), 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...

 

[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)?

 

[nobody56]

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

 

[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 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)?

 

[Dick]

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?

 

[Dick]

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?

 

[nobody56]

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

 

[Dick]

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

 

[nobody56]

right, so for b could I take the definition of a Centralizer C_G(H)={g element of G, such that ga=ag} and just plug in the elements to show C_G(H)=H? and do the same for the normalizer?

 

[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.

 

[nobody56]

So then would i just be able to say that since C_D4(r²)={1, r², s, sr²},and C_D4(s)={1, r², s, sr²}, and C_D4(sr²)={1, r², s, sr²}, then C_D4(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

 

[Dick]

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

 

[nobody56]

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