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## Putnam 2010 A5

Let $G$ be a group, with operation $\star$. Suppose that:

1. $G$ is a subset of $\Bbb{R}^3$;
2. For each $a,b \in G$ either $a \times b=a\star b$ or $a\times b=0$ (or both), where $\times$ is the usual cross product.

Prove that $a\times b=0$ for all $a,b \in G$.

Putnam A5

1. March 17, 2011 at 5:37 pm

I have a partial solution.

Let e be the identity element of G.
If this is the only element in G, we are done.

Suppose not.
Then there are two cases:
A) e is not 0
B) e = 0

In either case, choose any other element of G, say a.
Now either
1) a x e = a*e = a
or
2) a x e = 0

In case A):
If a is not zero, then a x e must be perpendicular to a.
So, if a x e = a, then a = 0

On the other hand, if a x e = 0, then a is some multiple of e

This if e is not 0, then every other element of G must be some multiple of e
(including 0).

But then a x b = 0 for all elements in G

The case e = 0 is giving me a lot more trouble.
Working on it.

• March 17, 2011 at 8:45 pm

Thank you. I will try and post a solution too. I think the case $e=0$ is harder than the other one.