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

In summary, the conversation is about proving that the centralizer of two elements in a group, a and gag^-1, is isomorphic. The centralizer of an element is the set of all elements in the group that commute with that element. The proposed mapping, f, is defined as f(h)=g^-1hg. The conversation also includes attempts at solving the problem and refining the solution.
  • #1
tyrannosaurus
37
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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:

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

What is the definition of isomorphic in abstract algebra?

The term "isomorphic" refers to a mathematical relationship between two algebraic structures, where there exists a one-to-one correspondence between their elements that preserves the structure and operations.

What are some examples of isomorphic structures in abstract algebra?

Examples of isomorphic structures in abstract algebra include groups, rings, fields, and vector spaces. For instance, the complex numbers are isomorphic to the real numbers, as they have the same algebraic properties and can be mapped to each other through a bijective function.

How do you prove that two structures are isomorphic?

To prove that two structures are isomorphic, you must demonstrate the existence of a bijective map between them that preserves the algebraic operations. This can be done by showing that the map is both injective (one-to-one) and surjective (onto).

What is the importance of isomorphism in abstract algebra?

Isomorphism is a fundamental concept in abstract algebra, as it allows us to study and understand different algebraic structures by relating them to each other. It also helps us identify and classify structures that have similar properties, making it a powerful tool in mathematical research.

Can two isomorphic structures be considered identical?

No, two isomorphic structures cannot be considered identical, as they may have different underlying elements and operations. Isomorphism simply means that there is a one-to-one correspondence between the structures, but they are still distinct entities.

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