# Putnam Problem 2010 A5

1. Dec 6, 2013

### stroustroup

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 $\mathbf{a}×\mathbf{b}=\mathbf{0}$.

Last edited: Dec 6, 2013
2. Dec 6, 2013

### haruspex

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