# The set of all conjugates of a and the set of all cosets of the Centralizer of a

1. Jun 30, 2012

### jmjlt88

I am working on constructing a bijection between the set of all conjugates of a and the set of all cosets of the Centralizer of a. Now, I let [a]={xεG: xax-1}. This is the set of all conjugates of a. The set {Cax : xεG} is the set of all cosets of Ca. Hence, I want a function f: [a] -> {Cax : x G}. I want to define f to be f(xax-1)= (xax-1)x. From a previous exercise, I am equipped with the fact that x-1ax= y-1ay if and only if Cax= Cay. Thus, if f(xax-1)=f(yay-1), then (xax-1)x= (yay-1)y which implies membership to both Cax and Cay. Hence xax-1= yay-1 and f is injective....

Before I go any further (i.e. prove that f is surjective and place QED at the end), is this the right idea? Or have I missed something or, perhaps, defined my function incorrectly? The fact that I am posting on here means I feel that something is amiss.

2. Jun 30, 2012

### micromass

Staff Emeritus
What do you mean with thus function. We have that $(xax^{-1})x$ is an element of G, it is not a coset of $C_a$. You need f to map $xax^{-1}$ to a coset of $C_a$.

3. Jun 30, 2012

### jmjlt88

Hmmm... What I am trying to do is map each element xax-1ε[a] to Cax for each xεG. Hence, define f(xax-1)=Cax..... Thus, if f(xax-1)=f(yay-1), then Cax= Cay which implies x-1ax= y-1ay..... AAAAAAAHHHHAAA!!! =) Better? I think it just hit me like a ton of bricks. You are the best micromass!

4. Jun 30, 2012

### micromass

Staff Emeritus
Be careful, you still want f to be well-defined. That is, if both $xax^{-1}$ and $yay^{-1}$ define the same element, then f of them is the same.