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Proof of uniqueness of square root

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Let G be a finite group in which every element has a square root. That is, for each x[tex]\epsilon G[/tex], there exists [tex]y \epsilon G[/tex] such that [tex]\(y^2=x.\)[/tex]Prove that every element in G has a unique square root.


    3. The attempt at a solution
    Proof: Assume not. Let k be the order of G. Let [tex]\(y_0, y_1, y_2,\text{...},\)\(y_k\)[/tex]be all the elements of G. For [tex]\(y_i{}^2\)= \(x_i\)[/tex]it can be said
    that [tex]\(y_i\)*\(y_i\)=\(x_i\)[/tex] and that [tex]\(y_i\)[/tex] is the square root of [tex]\(x_i\)[/tex]. If [tex]\(x_i\)[/tex] has more than one square root, then there would be two unique
    elements [tex]\(y_i\)[/tex] and [tex]\(y_j\)[/tex] with [tex]i \neq j[/tex] such that [tex]\(y_i\)*\(y_i\) = \(x_i\) =\(y_j\)*\(y_j\) = \(x_j\)[/tex]. However, if [tex]\(x_i\)[/tex] is the identity
    element, then [tex]\(y_i\)[/tex] must also be the identity element. However, [tex]\(x_i\) = \(y_j\)*\(y_j\)[/tex] and thus [tex]\(y_j\)[/tex] must also be the identity element. But
    [tex]\(y_i\)[/tex] and [tex]\(y_j\)[/tex] are distinct elements of G, and thus a contradiction is reached.



    I feel like I'm only proving that the identity element can not have two distinct square roots. Am I on the right path?
     
  2. jcsd
  3. Oct 10, 2008 #2
    Maybe this might be better: let y_i and y_j must have unique inverses. Let these be z_i and z_j respectively.
    y_i*y_i = x_i
    y_i*y_i = y_j * y_j
    y_i*y_i*z_i = y_j*y_j*z_i
    y_i = y_j*y_j*z_i
    z_i*y_i = z_i*y_j*y_j*z_i
    1 = z_i*y_j*y_j*z_i

    But this can only be true if z_i*y_j = 1 and y_j*z_i = 1, which can only occur if z_i and y_j are inverses. However, inverses are unique, and if z_i and y_i are inverses and z_i and y_j are inverses, then y_i must be y_j. However, this contradicts the original statement of y_i and y_j being distinct elements of G.
     
  4. Oct 10, 2008 #3

    Dick

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    Think of it this way. Let f(x)=x^2. f maps G to G. Since every element has a square root, f is onto. G is FINITE (very important). Can a map from a finite set G->G be onto if it's not one-to-one?
     
  5. Oct 10, 2008 #4
    Yeah, but I believe I'm supposed to be working only with the axioms of groups (associativity, inverse, identity element, law of exponents and the such).

    Edit: It's an Intro Abstract Algebra course.
     
  6. Oct 10, 2008 #5

    Dick

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    You also have to use the properties of a set being finite. Or it's not true. Take the group of complex numbers of magnitude 1 under multiplication. Every element has a square root. But the square root is not unique, every number has two.
     
  7. Oct 10, 2008 #6
    Ok, so if G has order k then...

    Let I be the set of all y_i 0=<i=<k
    and let J be the set of all x_i 0<i<k

    f(x) = x^2 maps I->J so that f(y_i) = x_i

    If some x_i has more than one square root, such as y_a and y_b, then there are only k-2 remaining elements in I to map to k-1 elements in J. So some element x_i could not have a square root unless y_c*y_c = x_c and x_d.

    Is this correct?
     
  8. Oct 10, 2008 #7

    Dick

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    Right. If one element has two square roots, then there are only k-2 candidates for the other k-1 elements that need square roots. There aren't enough to go around.
    I would stop talking after "So some element x_i could not have a square root". I'm not sure what the "unless" is about.
     
  9. Oct 10, 2008 #8
    Sorry, it's late. I was trying to say that unless some number squared could somehow have two values at the same time that it would be impossible for every element to have a square root. I'll go with what you said and just stop talking there.


    And now I feel stupid for not recognizing this earlier. Thanks for the help man!
     
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