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Exercise 15 from Hungerford: Algebra.

Let $G$ be a nonempty finite set with an associative binary operation such that for all $a,b,c\in G\,\,ab=ac \Rightarrow b=c$ and $ba=ca \Rightarrow b=c$. Then $G$ is a group. Show that this conclusion may be false if $G$ is infinite.

I've solved the first part, but I wasn't able to find a counter-example.

Thanks in advance!

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    After you are done solving the problem, the following reference has more: http://en.wikipedia.org/wiki/Cancellative_semigroup2012-01-24

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HINT: Is there a familiar (infinite) set and commutative operation in which you know that $ab=ac$ implies $b=c$? It's an example you can really count on.

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    Hello! Thanks a lot!2012-01-24
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    So that's what counter-example means!2012-01-24
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    It means that the cancellation law in a finite set with an associative binary operation cannot have a goup structure. Am I right? Sorry for the bad English!2012-01-24
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    @spohreis: Huh? No. If you have a binary associative operation on a finite set and the operation satisfies both cancellation laws, then the operation makes the set into a group; that's the point of the exercise. If you have a binary associative operation on *any* set (finite or infinite) that does *not* satisfy the cancellation laws, then the operation does not make the set into a group. In an **infinite** set, a binary associative operation that satisfies cancellation may or may not give you a group.2012-01-24
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    Sorry, I mean "infinite set".2012-01-24
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    @sopohreis: Still wrong: the integers under addition are an infinite set in which you have an associative binary operation which **does** give a group. The point is that for a binary associative operation, cancellation is *enough* in a finite set, but is *not enough* (though it is *necessary*) in an infinite set.2012-01-24
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    Ok! How about that? It is because the right (or the left) translation isn't bijective.2012-01-24
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    @spohreis: I don't understand what it is you are trying to do/say. What "is because" what? If you are trying to characterize what properties a binary associative operation on an infinite set must satisfy in order for it to give you a group, there are plenty of characterizations. E.g., that all equations $ax=b$ and $xa=b$ have solutions.2012-01-24
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    I am happy to be the vote which puts your rep in the 6 figures!! :) Congrats.2012-01-26
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    @SteveD: Thanks! But according to the audit I just ran, it's a bit premature. (-:2012-01-26