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Problem 15 of the first chapter, Groups, in Hungerford's Algebra states

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

It can be deduced from this that the inverse and the identity element are unique, but I can not understand how to attack the problem to prove $G$ may not be a group if it is infinite. How to approach the problem?

Orion 73
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  • What happens with the positive integers? – Arturo Magidin Nov 01 '20 at 05:20
  • Did you actually prove it is a group when $G$ is finite? You need to do more than just prove inverses and identity are unique if they exist; you also need to show they actually do exist. – Arturo Magidin Nov 01 '20 at 05:21
  • In case of positive integers, it is not a group both under multiplication and addition, since there's no inverse( in case of multiplication, viz inverse of 2 is 1/2 but 1/2 isn't an integer) and there's no identity element ( under addition, since 0 isn't a positive integer) – Orion 73 Nov 01 '20 at 05:32
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    But the point is that in the positive integers, which are infinite, under either addition or multiplication, the two conditions you have are satisfied. So that’s an example of a set with an associative operation which satisfies the conditions but is not a group. – Arturo Magidin Nov 01 '20 at 05:35
  • @ArturoMagidin Thank you I had missed the point. Is there any way to prove the statement without using counterexamples ? – Orion 73 Nov 01 '20 at 05:37
  • There are two statements to be proven. One of them requires a proof (that in the case of a finite set, the conditions do imply you have a group). The other (that when the set is infinite the conditions do not imply you have a group) requires a counterexample. You seem to be only asking about the latter. Are you also asking about the former? – Arturo Magidin Nov 01 '20 at 05:39
  • No, I'm asking about the latter, I've proven the former – Orion 73 Nov 01 '20 at 05:41
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    And what exactly do you think constitutes a proof that something need not happen which is not a counterexample? Obviously infinite groups exist, so you are not going to be able to show the conclusion does not hold in an infinite set. The way to show that “the conclusion may be false if $G$ is infinite” is to exhibit a $G$ that is infinite, satisfies the conditions, but is not a group. That is a counterexample. – Arturo Magidin Nov 01 '20 at 05:44
  • Thank you, I've got the point – Orion 73 Nov 01 '20 at 05:47
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    See also In a finite monoid, every left cancellable element is right invertible? – Bill Dubuque Nov 01 '20 at 06:30

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