Proving Equivalence Relations in Group Theory: Centralizer Math Problem

[DanielThrice]

(a) Let G be a group. Define ∼ by the following: a ∼ b ⇐⇒ ∃ g ∈ G such that gag^-1 = b.
Prove that ∼ is an equivalence relation.
(b) Suppose a ∈ Z(G). What elements are in the same cell as a with respect to the relation
∽?
(c) Let a ∈ G and define the centralizer of a, C_G(a), as the subset
C_G(a) = {g ∈ G : ga = ag}.
Prove that C_G(a) ≤ G. If a ∈ Z(G) what can you conclude about C_G(a)?
(d) (Bonus) Show that there is a bijective correspondence between elements equivalent (via
∼) to a ∈ G and left cosets of C_G(a).

I understand a). But I'm lost on the other three parts. This is not homework by the way per say, it was in my book and I'm studying for our final, this is one of the ones labeled "Very hard" basically. Thank you for your help

 

[Fredrik]

The rule is that all textbook-style questions should be treated as homework, even if they're not. So this should be in the homework forum, and you should show us your work so far. (I will request that this thread be moved to the homework forum). You should also explain what you mean by Z(G). By "cell", do you mean "equivalence class"?

 

[jtbell]

Thread moved to homework section.

 

[DanielThrice]

Well a is simple, I just proved that from a ~ b we can get b~a, and if c~ a and a ~ b, then c~b, thus ~ is an equivalence relation. And yes that is the correct definition of cell...and Z(G) is the center of G...this is the common notation for this.

 

[Fredrik]

In that case, b) is even easier than a). Just use the definitions of "cell", "center" and ~ (and of course "group"). In part c), does ≤ mean "is a subgroup of"? If yes, do you know how to check if a subset is a subgroup?

 

[DanielThrice]

Would this be correct for b?

Ok, commute means ga = ag.
What is gag-1?

b = gag-1 = agg-1 = ae = a

so only one element, a, can be in the same cell as a

And yea that means subgroup, and no I don't know how

 

[ECmathstudent]

To check if a subset is a subgroup you use the subgroup test theorem check to see if it's nonempty(usually the identity element), and then for closure and an inverse function.

 

[Fredrik]

Yes, that's what I had in mind for b).

Suppose that (G,m) is a group with underlying set G and multiplication operation m:G×G→G. A group (H,m') is said to be a subgroup of (G,m), if H is a subset of G and m' is the restriction of m to H×H.

If we know that (G,m) is a group, and that H is a subset of G, we can define m' to be the restriction of m to H×H. To prove that (H,m') is a subgroup, we need to show that

1. m'×H→H
2. m' is associative
3. e is in H.
4. For every x in H, x^-1 is in H. (Edit: I changed my typo x¹ to the correct x^-1 after the reply below...and corrected another typo in the word "associative").

It's convenient to use the notation xy both for m(x,y) and m'(x,y). This notation turns #1 into the following statement: If x and y are in H, the xy is in H. #2 is implied by the definition of m', and #3 and #4 are both implied by #1, so you only need to prove #1. (The first time you do this, you should also make the effort to understand why that is).

Edit 2: It's not true in general that #1 implies #3 and #4. That was a mistake on my part.

 

[DanielThrice]

H is CG(a) in this case. The operation m' is just * (written g*h or just gh).

We want to show * : CG(a) x CG(a) -> CG(a)
This means that for any g, h in CG(a)
We need to show gh or g * h is in CG(a)

g in CG(a) means what? ga = ag (commutative)
similarly ha = ah

How do we show gh is in CG(a)?
We need to show (gh)a = a(gh)
How? (gh)a ={associative}= g(ha) ={h in CG(a)}= g(ah) ={assoc}= (ga)h ={g in CG(a)}= (ag)h ={assoc}= a(gh)

We can assume associativity (see below)

This proves 1



Now 2, associativity, is trivial, because any g,h,k in CG(a) is also in G which is already associative by being a group.

For 3, what is e? How do we show e is in CG(a)?

For 4, what is x^{-1}? If x is in CG(a), how can we show x^{-1} is in CG(a)?

I'm still stuck on those two...

 

[Fredrik]

H is CG(a) in this case. The operation m' is just * (written g*h or just gh).

We want to show * : CG(a) x CG(a) -> CG(a)
This means that for any g, h in CG(a)
We need to show gh or g * h is in CG(a)

g in CG(a) means what? ga = ag (commutative)
similarly ha = ah

How do we show gh is in CG(a)?
We need to show (gh)a = a(gh)
How? (gh)a ={associative}= g(ha) ={h in CG(a)}= g(ah) ={assoc}= (ga)h ={g in CG(a)}= (ag)h ={assoc}= a(gh)

We can assume associativity (see below)

This proves 1
Now 2, associativity, is trivial, because any g,h,k in CG(a) is also in G which is already associative by being a group.

Perfect.

For 3, what is e? How do we show e is in CG(a)?

e is the identity element of the group G. It's in C_G(a) because ea=ae (=e).
Edit: I said something incorrect here, so I deleted it.

For 4, what is x^{-1}? If x is in CG(a), how can we show x^{-1} is in CG(a)?

I'm still stuck on those two...

x^-1 is the inverse element of x. I seem to be having some kind of brain malfunction...right now I don't see it either. I have to go out for a bit, but I'll have another look at it when I come back.

 

[Fredrik]

OK, I actually thought that 1 would imply both 3 and 4, but that seems to be false. So we have to prove both 3 and 4 for this special case. I did 3 in my previous post. 4 is easy too: Suppose that x is in C_G(a). Then x^-1a=x^-1a(xx^-1)=... and I'm sure you can figure out the rest.