# An equivalent definition of a group

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: en.wikipedia.org/wiki/Cancellative_semigroup – Jonas Meyer Jan 24 '12 at 20:08

## 1 Answer

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! – spohreis Jan 24 '12 at 20:03
So that's what counter-example means! – hardmath Jan 24 '12 at 20:12
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! – spohreis Jan 24 '12 at 20:15
@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. – Arturo Magidin Jan 24 '12 at 20:20
@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. – Arturo Magidin Jan 24 '12 at 20:38