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Let G be a finite group in which every element has a square root

  1. Mar 8, 2013 #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 in G, there exists a y in G such that y^2=x. Prove every element in G has a unique square root.

    2. Relevant equations

    G being a group means it is a set with operation * satisfying:
    1.) for all a,b,c in G, a*(b*c)=(a*b)*c
    2.) there exists an e s.t. for all x in G, x*e=x
    3.) for all x in G there exists an x' s.t. x'*x=e

    3. The attempt at a solution

    The only thing I have gotten so far is the reason why (-2)^2 and 2^2 both equaling 4 doesn't present a counterexample. The reason it doesn't is because (-2) has no square root so it is not in our group G.

    ... other than that I'm stuck..
     
  2. jcsd
  3. Mar 8, 2013 #2

    Dick

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    You are probably overthinking this. If you have two different elements of the group, x1 and x2 and y1 is a square root of x1 and y2 is a square root of x2, then y1 is not equal to y2, yes? So you have at least as many square roots in G as there are elements of G. What happens if an element has two square roots? Use that G is finite.
     
    Last edited: Mar 8, 2013
  4. Mar 8, 2013 #3
    Every element in G has a square root. If an element has 2 square roots then two elements in G have the same square root (by the pidgeonhole principle). That means there exists a y in G s.t. y^2=x1 and y^2=x2 where x1=/=x2. This is a contradiction because it would suggest y^2=/=y^2.


    Is this right?

    Thanks for your help
     
  5. Mar 8, 2013 #4

    Dick

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    You've got the idea, but I guess I would start with proving all of square roots are distinct elements of G first, instead of leaving it for a contradiction at the end. And why does G have to be finite?
     
  6. Mar 8, 2013 #5
    I don't get what you mean by this. Do you mean prove there is no element y such that y is the square root of x1 and x2, x1=/=x2? Or do you mean prove that for any x, the "square root of x" is well-defined?

    I don't see what's wrong about what I did? I assumed BWOC that there existed an element with 2 different square roots and I showed how that led to a contradiction... Doesn't that imply no element has 2 square roots? And thus every element has a unique square root?


    Thanks


    If G was infinite we couldn't use the pidgeonhole principle?
     
  7. Mar 8, 2013 #6

    Dick

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    I mean prove there is no element y such that y is the square root of x1 and x2, x1=/=x2. And yes, I know it's obvious. "If an element has 2 square roots then two elements in G have the same square root", that doesn't ring right. In the integers mod 3, 1^1=1 and 2^2=1. So the element 1 has two square roots. That doesn't mean two elements have the same square root. At least until you explain it better. And, yes, I know 2 doesn't have a square root.

    I'm not saying it's wrong. Just explain what the pigeons are and what the holes are.
     
    Last edited: Mar 8, 2013
  8. Mar 19, 2013 #7
    I should have said: "If an element has 2 square roots then two elements in G share a square root."

    The reason I feel that's true is because since G is finite, there are only |G| possible options for an element's square root. And there are also |G| elements each of which has a square root. So there are |G| pidgeons (being the elements of G), and |G| holes (the possible square roots). Therefore if one pidgeon has 2 square roots, then that leaves |G|-2 pidgeonholes for |G|-1 remaining pidgeons... and thus two elements must share a square root.

    x1 and x2 sharing a square root means there is a y in G such that y^2=x1 and y^2=x2... but that's a contradiction?


    Is that right?
     
  9. Mar 19, 2013 #8

    Dick

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    Yes, it sounds correct.
     
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