## Putnam 2010 A5

Let be a group, with operation . Suppose that:

- is a subset of ;
- For each either or (or both), where is the usual cross product.

Prove that for all .

*Putnam A5*

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Categories: Higher Algebra, Problem Solving
putnam

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.

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