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Group set proof

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Let (G,*) be a finite group of even order. Prove that there exists some g in G such that g≠e and g*g=e. [where e is the identity for (G,*)]


    2. Relevant equations
    Group properties


    3. The attempt at a solution
    Let S = G - {e}. Then S is of odd order, and let T={g,g^-1: g[itex]\in[/itex]S}.
    Then [itex]\exists[/itex]h[itex]\in[/itex]S such that h[itex]\notin[/itex]T. Since G is a group, h*h must equal e.

    Does that work?
     
  2. jcsd
  3. Dec 6, 2012 #2
    You have a good idea here, nearly perfect, but the last step seems not obvious at first. Who says that h*h must be equal to e? All we know (since G is a group) is that h*h must lie in G. To show that h*h = e, you also need h∉S. (Then h*h ∈ G - S = {e}.)
     
  4. Dec 6, 2012 #3

    jbunniii

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    How is that possible? You have defined T in such a way that [itex]S \subset T[/itex]. You have the right idea, though. If you partition G such that each element of g is partnered with its inverse, then e will be partnered with itself. If you assume that no other element has this property, then you should be able to obtain a contradiction.
     
  5. Dec 6, 2012 #4
    If h[itex]\notin[/itex]S, then h=e, which isn't what we want. I was trying to show that if you take all the elements and their inverses out of S, then you're left with a single element whose inverse must be itself. There's probably a better way to show this.
     
  6. Dec 6, 2012 #5
    How about collecting all elements that aren't their own inverses in one set, i.e. S = { g∈G : g-1≠g }. Then for any element x∈G, either x ∈ G - S, or x and x-1 are both in S. "Both" is important here, since it means that S has an even number of elements.

    So what's left in G - S?
     
  7. Dec 6, 2012 #6
    That's good, thanks!
     
  8. Dec 6, 2012 #7

    micromass

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    It might be interesting for the OP that this result has a significant generalization: If G is a group and if p is a prime number such that p divides the order of G, then there exists an element g in G with order p.

    This is called Cauchy's theorem.
     
  9. Dec 6, 2012 #8
    So |x|[itex]\in[/itex]G = p? Would this relate to a group being cyclic?
     
  10. Dec 6, 2012 #9

    micromass

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    I don't really get your notation of [itex]|x|\in G=p[/itex] :confused:

    I mean to say that there exists a [itex]x\in G[/itex] such that [itex]|x|=p[/itex] (assuming |x| means order for you).

    If G has prime order, then this result shows indeed that there exists an element in G with prime order. So G must be cyclic. Is this a relationship that you are talking about?
     
  11. Dec 6, 2012 #10
    Yeah that's exactly it, I was being sloppy. My understanding of a cyclic group is that there exists some element of G, say a, such that, for every x in G, x=a^n for some integer n.

    How does this follow from |x|=p?
     
  12. Dec 6, 2012 #11

    micromass

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    It doesn't follow in general.

    I'm saying the following: If G has order p, then there exists an element a in G that has order p. So [itex]\{1,a,a^2,...,a^{p-1}\}[/itex] has p elements and is a subset of G. Since G also has p elements, we must have [itex]G=\{1,a,a^2,...,a^{p-1}\}[/itex]. So G is cyclic.
     
  13. Dec 7, 2012 #12
    Look at all elements of the form xn with n=0,...,(p-1). If you can show that they're all distinct (i.e. xn ≠ xm if n ≠ m), then these p elements must make up all of G.
     
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