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Proving a function is a bijection and isomorphic

  1. Mar 28, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    If G is a group and a ∈ G, let π: G--> G be the function defined as π(g) = ag, for all g ∈G.
    a) Show that π is a bijection
    b) Show that if π is an isomorphism, then a is the identity element of G.

    2. Relevant equations
    I think to show that pi is a bijection we have to show that it is surjective and injective.
    To show that it is an isomorphism we have to show that π(xy) = π(x)π(y).

    3. The attempt at a solution

    First, I do part a and show that it is one to one:
    Suppose π(g) = π(g')
    then ag = ag'
    Since a ∈G we know a^(-1) exists in G.
    a^(-1)ag = a^(-1)ag'
    g = g'
    and so π is one to one.

    I am not sure how to prove that this is onto... any hints here would help.
    For onto can I just say
    Let x,y ∈ G and x = a^(-1)y
    since a is a part of the group we know that a^(-1) exists.
    And thus it follows that π(x) = y

    To show part b)
    if x,y ∈G, then we observe
    π(xy) = axy
    π(x)π(y) = axay

    Here, a must equal the identity element of G for this function to be an isomorphism.
     
  2. jcsd
  3. Mar 28, 2016 #2

    Samy_A

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    Mainly correct.

    Only the last point needs some explanation.
    You have that, for π to be an isomorphism, axy = axay, for all x,y ∈ G.
    Why does that imply that a is the identity element of G?
     
  4. Mar 28, 2016 #3

    RJLiberator

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    Hm, well, it implies that a is the identity element of G as it would mean that a=aa
    the only thing that works here is the identity element.
     
  5. Mar 28, 2016 #4

    RJLiberator

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    Maybe you want it more explicit?

    axy = axay

    y^(-1) exists as y is a part of G so multiple both sides by it

    axy*y^(-1) = axay*y^(-1)

    ax = axa

    a^(-1) exists as a is a part of G
    a^(-1)ax = a^(-1)axa
    x = xa
    so now a has to be the identity element
     
  6. Mar 28, 2016 #5

    Samy_A

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    Voila!
     
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