The post that was quoted here has been removedNot 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.
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The post that was quoted here has been removedA 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 SoothfastI 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).
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.
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.
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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.
The post that was quoted here has been removedIt'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.