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Isomorphism is an equivalence relation on groups

  1. Mar 12, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Prove that isomorphism is an equivalence relation on groups.

    2. Relevant equations

    Need to prove reflexivity, symmetry, and transitivity for equivalence relationship to be upheld.

    **We will use ≅ to define isomorphic to**
    3. The attempt at a solution

    Let G, H, and K be groups.
    We want to show that G≅G, G≅H, and if G≅H H≅K then G≅K.

    Reflexivity: Define f:G->G by f(x)=x. we show f is an isomorphism. For x,y ∈ G if f(x) = f(y), then by definition of f we see that x=y. f is thus one to one. f(x)=x so f is onto. f(xy) = xy = f(x)f(y).
    There f is an isomorphism and G≅G.

    Symmetric: We assume G≅H. Then there is an isomorphism f:G->H, that is a bijection and homomorphism.
    Claim: f^(-1) from H to G is an isomorphism.
    Since f is a bijection, so is f^(-1).
    let h, h' ∈H and g,g' ∈G. f^(-1)(h*h') = f^(-1)(h)f^(-1)(h').
    Since f^(-1) is a bijection, there exists g,g'∈G such that f(g) = h and f(g') = h'.
    f(gg') = f(g)f(g') = hh' => f^(-1)(hh') = gg'.
    f^(-1)(h)f^(-1)(h') = gg'.
    Thus, f^(-1)(hh') = f^(-1)(h)f^(-1)(h') as required. Therefore H≅G.

    Transitive: Assume G≅H and H≅K, then there are isomorphisms g:G->H and f:H->K.
    Claim: (g°f):G->K is an isomorphism.
    The composition of two bijections is also a bijection.
    For any h,h' ∈G. (g°f)(hh') = g(f(hh')) = g(f(h)f(h')) = g(f(h))g(f(h')) = (g°f)(h)(g°f)(h')
    Hence, G≅K.

    Thus, isomorphisms of groups is an Equivalence relationship.



    Question: I am most concerned about the symmetric part. Did I do this part right?

    How can I arbitrarily define isomorphisms like f:G->G by f(x)=x?

    Thank you.
     
  2. jcsd
  3. Mar 12, 2016 #2

    PeroK

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    It looks okay to me. I've underlined one thing that I think is superfluous. On the last question, you can say "let I or f be the identity mapping".
     
  4. Mar 12, 2016 #3

    RJLiberator

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    Is the underlined part due to me using an asterisk there when it is not needed?
     
  5. Mar 12, 2016 #4

    PeroK

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    The bit I underlined can all be deleted. It's not necessary. Nothing to do with the asterisk.
     
  6. Mar 12, 2016 #5

    RJLiberator

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    I see. Reading it through, that makes sense. Thanks.
     
  7. Mar 12, 2016 #6
    Yes if ##f## is a group isomorphism, so is its inverse. However, you don't explicitly assign internal laws in your notations, and it can be bothering for the reader.
     
  8. Mar 12, 2016 #7

    RJLiberator

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    I am unfamiliar with this term 'internal laws'
     
  9. Mar 12, 2016 #8
    When you describe a group, you have to specify a pair ##(S,\star)##, where ##S## is a set, and ##\star : S\times S \to S ## is the composition law ( or internal law ).
     
  10. Mar 12, 2016 #9

    RJLiberator

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    Ahh, I get it. So you are saying, perhaps I should assign internal laws to my proofs for easier reading/ a more clear proof. Such as (S,*).
     
  11. Mar 12, 2016 #10
    Yes, just for easier reading, not to criticize you
     
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