Is C(a) isomorphic to C(gag-1) for elements a and g in a group?

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Homework Help Overview

The discussion revolves around the isomorphism between the centralizer of an element \( a \) and the centralizer of the conjugate \( gag^{-1} \) in a group. Participants are exploring the properties of a proposed mapping between these two centralizers.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss defining a mapping \( f \) from \( C(gag^{-1}) \) to \( C(a) \) and question whether this mapping is well-defined. There are attempts to establish properties such as injectivity and surjectivity, along with operation preservation.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts. Some express uncertainty about the completeness of the justification for the mapping's properties, while others seek clarification on the definitions and implications of isomorphism.

Contextual Notes

There is an emphasis on ensuring that the mapping is well-defined and that it satisfies the necessary properties to establish an isomorphism. Participants are also considering the expectations of their professor regarding the rigor of their arguments.

tyrannosaurus
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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.
 
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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.
 
tyrannosaurus said:
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.
 
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.
 
Last edited:

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