# Abstract algebra isomorphic

## Homework Statement

If a and g are elements of a group, prove that C(a) is isomorphic to C(gag-1)

## Homework Equations

I have defined to mapping to be f:C(gag-1) to C(a) with f(h)=g-1hg.
I have no idea if this is right.

## The Attempt at a Solution

I don't have a clue at the solution, any help would be greatly appreciated.

forgot to mention, that C(a) and C(gag^-1) means the centralizer of a and the centralizer of gag^-1. The centralizer of a is the set of all elements in G that commute with a.

Hurkyl
Staff Emeritus
Gold Member
I have no idea if this is right.
Have you tried proving it's an isomorphism? Or at least stating what that would mean?

here is my new attempt at the solution:
1. Mapping: F:C(gag^-1) to C(a) is obviously a well defined function with f(h)=g^-1hg.
2. One to one- Let h and l be elements of C(gag^-1). Then by definition of f, g^-1hg=g^-1lg, this h=l by left and right cancellation laws.
3. Onto- let k=gmg^-1 that is an element of C(a). Let p=m. Obviously p is an element of C(gag^-1). Then f(p)=f(m)=gmg^-1. Thus f is onto.
4. Operation oreservation
Let r,s be elements of C(gag^-1). Then f(r*s)=(g^-1)*r*s*g= (g^-1)*r*e*s*g *(e=identity element)=(g^-1)*r*g*(g^-1)*s*g=((g^-1)*r*s)*((g^-1)*s*g)(associativity of operation)=f(r)*f(s).
Thus f preserves the operation and C(a) and C(gag^-1) are isomorphic.
Does this sound any better.

Hurkyl
Staff Emeritus
Gold Member
My issues:

1. To be a well-defined function C(gag-1) -> C(a), the relation you defined
f(h) = g-1hg​
has to have two properties:
• It must be a function
• C(a) contains the image of C(gag-1)
I will agree that it's obviously one-to-one. I suspect your professor would prefer more justification on the second part.
2. This is more-or-less fine. I think you forgot a phrase like "such that f(h)=f(l)" somewhere, though.
3. I'm really confused about what you're doing here. (Also, I suspect your professor would prefer "obviously" to be replaced with something more details)
4. This is good.

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