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Putnam Problem 2010 A5

  1. Dec 6, 2013 #1
    I am trying to solve past Putnam Competition problems. I think I have a proof for Problem A5 of the 2010 exam. Here is the set of all problems: http://inside.mines.edu/~dlarue/putnam/exams/2010.pdf

    The solution given required three lemmas to prove, so I wonder whether my proof is flawed (it seems too simple, and the only group property I used was existence of a (right) identity). So can you tell me if my proof is correct?

    Let [itex]\mathbf{e}[/itex] be an identity of the group [itex]G[/itex]. Suppose there exists a vector [itex]\mathbf{a}[/itex] in [itex]G[/itex] such that [itex]\mathbf{a}×\mathbf{e}≠\mathbf{0}[/itex].
    Then [itex]\mathbf{a}×\mathbf{e} = \mathbf{a}\ast\mathbf{e}=\mathbf{a}[/itex], therefore [itex]\mathbf{a}×\left(\mathbf{a}×\mathbf{e}\right) = \mathbf{a}×\mathbf{a} = \mathbf{0}[/itex].
    But by properties of the cross product, [itex]\|\mathbf{a}×\left(\mathbf{a}×\mathbf{e}\right)\|=\| \mathbf{a}\|\| \mathbf{a}×\mathbf{e}\| ≠ \mathbf{0}[/itex] since neither [itex]\mathbf{a}[/itex] nor [itex]\mathbf{a}×\mathbf{e}[/itex] are [itex]\mathbf{0}[/itex].

    The obtained contradiction shows that [itex]\mathbf{a}×\mathbf{e} = \mathbf{0}[/itex] for all [itex]\mathbf{a}[/itex] in [itex]G[/itex]. Therefore, every vector in [itex]G[/itex] is proportional to the vector [itex]\mathbf{e}[/itex]. But this means that all vectors lie on the same line, and therefore for every [itex]\mathbf{a},\mathbf{b}[/itex] in [itex]G[/itex], we have [itex]\mathbf{a}×\mathbf{b}=\mathbf{0}[/itex].
     
    Last edited: Dec 6, 2013
  2. jcsd
  3. Dec 6, 2013 #2

    haruspex

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    Your proof looks sound to me (and very neat).
     
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