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Homework Help: Abstract algebra isomorphic

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant 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.

    3. The attempt at a solution
    I don't have a clue at the solution, any help would be greatly appreciated.
     
  2. jcsd
  3. Feb 23, 2010 #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.
     
  4. Feb 23, 2010 #3

    Hurkyl

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    Have you tried proving it's an isomorphism? Or at least stating what that would mean?
     
  5. Feb 23, 2010 #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.
     
  6. Feb 24, 2010 #5

    Hurkyl

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    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: Feb 24, 2010
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