Easy problem

Well, it's easy if you spot the trick. Suppose a set S has a product which is anti-commutative. So a*b = -b*a, show that it is not associative. You can assume that there is at least one product where a*b != a*c (!= means not equal to).

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Not every set has a defined product, but the OP states that there is a product defined in the set. He assumes you know what 'product' means in this instance. He has not said that the set in question is a group under any given operation, and I assume it is not a requirement of the problem.

If you want to talk about a 'product', then you must define an operation upon elements of the set.
Yes, that is what he did.

Originally posted by DeepThought
Well, it's easy if you spot the trick. Suppose a set S has a product which is anti-commutative. So a*b = -b*a, show that it is not associative. You can assume that there is at least one product where a*b != a*c (!= means not equal to).

Edited

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This is a chess site.

And I don't think the fun question is being submitted for a finals paper.

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A group is necessarily associative, so that is why DeepThought doesn't give that the set is a group.

EDIT: While I have not given any thought to DeepThought's riddle, there is a fair chance his solution runs on the assumption that the set S is "closed" under the product. That is, if a,b ∈ S, then a*b ∈ S.

Originally posted by Soothfast
A group is necessarily associative, so that is why DeepThought doesn't give that the set is a group.

EDIT: While I have not given any thought to DeepThought's riddle, there is a fair chance his solution runs on the assumption that the set S is "closed" under the product. That is, if a,b ∈ S, then a*b ∈ S.

I think technically it's a module since I've implicitly got addition defined (otherwise -a*b doesn't make sense). I didn't want to give any more information in the question than was necessary. The stuff about at least one case where a*b != a*c was to rule out a trivial case (where a*b = 0 for all elements in the set, when it is trivially associative).

Yes I think I'd better make it closed under the operation (as I'm not certain of the effects of not doing that, and certainly the specific example I generalised from has that property) - although I don't think that that is necessary for the answer.

Drat, didn't think it through enough I need to add two extra conditions. It's best if I restate the problem:

Let S be a set which is closed under the product * and has the following properties:

Anti-commutivity a*b = -b*a
If a and b are different elements then a*b != 0
a*b != a and a*b !=b for all a and b.

Show that S is not associative under '*'

Sorry, I tried to abstract away from a specific example and worried about trivial cases instead of whether I'd given enough information. The extra two conditions may overspecify it, but I wanted to make it different to the specific case I based it on to prevent people just looking the answer up on Wikipedia.

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It's a module under addition and multiplication by a scalar, with a product operation which is anti-commutative. In the restated problem - I didn't give enough information - I'm not relying on the multiplication by a scalar so it need only be an abelian group under addition.

The problem is now easy, where before hand it's probably not possible.