
Centralizer
a) Let G be a group. Deﬁne ∼ by the following: a ∼ b ⇐⇒ ∃ g ∈ G such that gag1 = 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, CG(a), as the subset
CG(a) = {g ∈ G : ga = ag}.
Prove that CG(a) ≤ G. If a ∈ Z(G) what can you conclude about CG(a)?
(d) (Bonus) Show that there is a bijective correspondence between elements equivalent (via
∼) to a ∈ G and left cosets of CG(a).
I understand a), I proved ~ is an equivalence relation. 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
Just for clarity, Z(G) indicates the center of the group

For part (b), what does center mean? a commutes with every other element of G.
If a commutes what is gag1?
So if b ~ a then b = gag1 = ...

That's not really progressing anything...and yes the center is everything that commutes with everything else, think of it as the abelian portion of the group

For part (c) you need to prove that CG(a) is a subgroup.
That is show
closure: g, h are in CG(a), then is gh in CG(a)?
ga = ag
ha = ah
show gha = agh
inverse: g is in CG(a) is g1 in CG(a)?
ga = ag
show g1a=ag1
(hint g1a = g1agg1)
Also, when a is in Z(G) (a commutes with every element of G), what is CG(a) (all elements of G that commute with a)?

Ok, commute means ga = ag.
What is gag1?
b = gag1 = agg1 = ae = a
so only one element, a, can be in the same cell as a

Wouldn't that equality mean that b is also in the cell, or does that mean that any other eloement, to be in the cell, has to be equal to a, and thus a is the only one in the cell?

b is only notation for an arbitrary element.
We are saying, consider an arbitrary element b in the cell of a (could be anything we want), and we proved b = a (always no matter how we picked b). This means that any element in the cell of a is equal to a.
Hence, a is the only element in the cell. gag1 cannot be anything but a.

Alright, that makes sense, thank you, but as for c i'm still unsure about it, it's all elements of G that commute with just a?

For (c), CG(a) is the *subset* of G of all elements that commute with a.
But now you have to show that this *set* is a *subgroup* of G, CG(a) <= G
This amounts to using the definitions of subgroup, which I have highlighted in an earlier post.

Alright, so I guess 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→H
2. m' is asociative
3. e is in H.
4. For every x in H, x1 is in H.
But how exactly do I get it down from here

You already have everything. Just disect the definitions.
H is CG(a) in this case. The operation m' is just * (written g*h or just gh).
I will disect 1 in detail for you:
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)?
Already provided the hint in above post (this one is a bit tricky).
After 4 you are done :D

Alrighty, so e is the identity element of the group G. It's in CG(a) because ea = ae (=e) right?
As for the inverse, how does this compare to what you did and is it legal...
We can assume that x1a = x1a
x1a = x1a
x1a = x1ae
x1a = x1a(xx1)
x1a = ax1(xx1)
x1a = a(x1x)x1
x1a = aex1
x1a = ax1

Yup, you got it.
Just remember, for algebra problems, keep expanding definitions, and you will be half way to the solution.