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Proving a function is an isomorphism

  1. Mar 26, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Let G be a finite abelian group with no elements of order 2 Show that the function φ: G-> G defined as φ(g) = g^2 for all g ∈G, is an isomorphism.


    2. Relevant equations
    Abelian group means xy = yx for all x,y∈G

    Isomorphic if there exists a bijection ϒ: G_1 -> G_2 such that for all x,y ∈ G, ϒ(xy) = ϒ(x)ϒ(y)

    3. The attempt at a solution

    We have a couple of main points.
    -Abelian
    -Definition of the function maps one element to that element squared.
    -G is finite with no elements of order 2.

    To prove that it is isomorphic, we use the definition, that there must exist a bijection from G1 to G1 such that for all xy we see ϒ(xy) = ϒ(x)ϒ(y)

    I'm not sure where to go from here. Do I just say for some x, y in G we have ϒ(xy) = ϒ(yx) = (xy)^2 = (yx)^2 or some form of this ?
     
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  3. Mar 27, 2016 #2

    Samy_A

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    Well, a group is obviously isomorphic with itself.
    What you have to prove is that φ is a group isomorphism.

    First thing you have to prove is that φ is a group homomorphism. You more or less did that, although I'm not sure why you used Y and not φ.

    Once it is proved that φ is a group homomorphism, you have to prove it is a bijection.
     
    Last edited: Mar 27, 2016
  4. Mar 27, 2016 #3

    RJLiberator

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    Thanks for the help.

    How is this looking:

    We seek to show that φ is a group isomorphism. First we show that φ is a group homomorphism.

    φ(xy) = φ(yx) = (xy)^2 = (yx)^2
    (xy)^2 = x^2y^2 = φ(x)φ(y)

    Therefore φ(xy) = φ(x)φ(y) and we've proven that it is a group homomorphism.

    To the prove that φ is a bijection we show the following:

    Given a fixed g in G,
    g*x = g*y

    g^(-1) * g*x = g^(-1) * g*y
    x = y.

    so we see one to one correspondence.
    Now,
    Let x = g^(-1)*y
    g*x = y
    so φ is onto.
     
  5. Mar 27, 2016 #4

    Samy_A

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    Correct.
    I'm not sure I understand what you do here. What is that fixed g in G?

    You have to show that φ(x)=φ(y) implies x=y.
    Again, what exactly is g?

    You have to show that, given any y ∈ G, ∃ x ∈ G such that φ(x)=y.
     
  6. Mar 27, 2016 #5

    RJLiberator

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    Could you just say:

    Assume φ(x) = φ(y)
    then x^2 = y^2 and so x must equal y.

    And then
    if we let x = φ(-1)*y then φ(x) = φ(φ^(-1)y) = y
     
  7. Mar 27, 2016 #6

    Samy_A

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    You have to explain why x²=y² implies x=y. That is not true in general in a group (in ℤ, (-1)²=1², for example).
    What is φ(-1)?
    You have to show that φ is a bijection, so you can just assume φ-1 exists (assuming that's what you meant with φ(-1)).
     
  8. Mar 27, 2016 #7

    RJLiberator

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    x^2 = y^2 implies x = y as the group is abelian no elements of order 2.
    Correct?

    Sorry, yes, that is what I meant with φ^(-1).
     
  9. Mar 27, 2016 #8

    Samy_A

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    Yes, that is correct, but somewhat terse.
    An examinator might expect a more detailed argument of how the group being Abelian and not having elements of order 2 implies that x²=y² means that x=y.
    So how do you prove that φ is surjective (onto)?
     
  10. Mar 27, 2016 #9

    RJLiberator

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    I assume that φ^(-1) exists so we let x = φ^(-1)*y

    We see φ(x) = φ(φ^(-1)*y) = y.
    So φ is onto.

    Absolutely. Hm.
    I'm having trouble seeing it. By no elements of order 2, and since it is abelian, we know that it either has elements of order greater than 2 or just of order 1.
     
  11. Mar 27, 2016 #10

    Samy_A

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    You can't assume that φ-1 exists, that is part of what you have to prove.

    You have to do it the "hard" way.
    Take y ∈ G. You need to find an x ∈ G such that φ(x) = x² = y.

    Hint: since G is finite, y as an order, ie ym = e for some integer m (e is the unit element of G). Is m even or odd?
    Yes, but that doesn't help much, does it?
    Hint: assume you have x² = y². Since x and y are elements of the group G, they have inverses, x-1, y-1.
     
    Last edited: Mar 27, 2016
  12. Mar 27, 2016 #11

    RJLiberator

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    Take y ∈ G. You need to find an x ∈ G such that φ(x) = y.
    Could you say that since y^m = e (as discussed), then φ(x) = y would be (φ(x))^m = e and so x = e ?


    For the last part, since x and y are elements of the group G they have inverses x^(-1) and y^(-1).
    So since we arrived at x^2 = y^2 we can then take x^(-1) on both sides and see
    x^(-1)*x*x = x
    and x^(-1) *y*y on the other side.

    so x = x^(-1)*y*y

    now take y^(-1) and see
    y^(-1)*x = x^(-1)*y

    Not sure if this helps tho...
     
  13. Mar 27, 2016 #12

    Samy_A

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    Not really. You don't have x yet.

    Let's do it in steps. If m is the order of y, is m an even or an odd integer?

    Way too complicated. Why not multiply both sides twice by y-1, and remember that the group is abelian?
     
  14. Mar 27, 2016 #13

    RJLiberator

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    Ah... yes....

    It would be odd as there are no elements of order 2.
    EDIT: wait, since the map takes an element in g and maps it to that element ^2 we see that it is even.
    EDIT:EDIT My edit was wrong :P
     
  15. Mar 27, 2016 #14

    Samy_A

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    Correct.
    So, ym = e, with m some odd integer.
    Then what is ym+1? And can this lead to an x ∈ G satisfying φ(x) = x² = y?
     
  16. Mar 27, 2016 #15

    RJLiberator

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    y^(m+1) = y?

    So then if some x = y^((m+1)/2) then x^2 would equal y?
     
  17. Mar 27, 2016 #16

    Samy_A

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    Yes, x = y(m+1)/2 satisfies φ(x) = y. That proves that φ is surjective.
     
  18. Mar 27, 2016 #17

    RJLiberator

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    wooooohoooo. What beauty.
     
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