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stroustroup
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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 \mathbf{e} be an identity of the group G. Suppose there exists a vector \mathbf{a} in G such that \mathbf{a}×\mathbf{e}≠\mathbf{0}.
Then \mathbf{a}×\mathbf{e} = \mathbf{a}\ast\mathbf{e}=\mathbf{a}, therefore \mathbf{a}×\left(\mathbf{a}×\mathbf{e}\right) = \mathbf{a}×\mathbf{a} = \mathbf{0}.
But by properties of the cross product, \|\mathbf{a}×\left(\mathbf{a}×\mathbf{e}\right)\|=\| \mathbf{a}\|\| \mathbf{a}×\mathbf{e}\| ≠ \mathbf{0} since neither \mathbf{a} nor \mathbf{a}×\mathbf{e} are \mathbf{0}.
The obtained contradiction shows that \mathbf{a}×\mathbf{e} = \mathbf{0} for all \mathbf{a} in G. Therefore, every vector in G is proportional to the vector \mathbf{e}. But this means that all vectors lie on the same line, and therefore for every \mathbf{a},\mathbf{b} in G, we have \mathbf{a}×\mathbf{b}=\mathbf{0}.
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 \mathbf{e} be an identity of the group G. Suppose there exists a vector \mathbf{a} in G such that \mathbf{a}×\mathbf{e}≠\mathbf{0}.
Then \mathbf{a}×\mathbf{e} = \mathbf{a}\ast\mathbf{e}=\mathbf{a}, therefore \mathbf{a}×\left(\mathbf{a}×\mathbf{e}\right) = \mathbf{a}×\mathbf{a} = \mathbf{0}.
But by properties of the cross product, \|\mathbf{a}×\left(\mathbf{a}×\mathbf{e}\right)\|=\| \mathbf{a}\|\| \mathbf{a}×\mathbf{e}\| ≠ \mathbf{0} since neither \mathbf{a} nor \mathbf{a}×\mathbf{e} are \mathbf{0}.
The obtained contradiction shows that \mathbf{a}×\mathbf{e} = \mathbf{0} for all \mathbf{a} in G. Therefore, every vector in G is proportional to the vector \mathbf{e}. But this means that all vectors lie on the same line, and therefore for every \mathbf{a},\mathbf{b} in G, we have \mathbf{a}×\mathbf{b}=\mathbf{0}.
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