# A set which satisfies all conditions for a Group except associativity

I have a question which I can't seem to prove or dismiss.

Can a set of elements A satisfy al the conditions for a group except associativity (which leaves us with closure, identity and invertibility).

Tried to prove it but can't seem to make a table which fullfiles these actions.

Thank you.

• Subtraction.${}$ – user1729 Jul 28 '13 at 17:59
• Since there are different sets of axioms of groups (which are equivalent as long as associativity is given, but not without associativity) you should make precise what you mean with "all" the conditions. – Hagen von Eitzen Jul 28 '13 at 18:17
• Just make the product of any two nonidentity elements equal the identity. – MartianInvader Jul 28 '13 at 19:58
• Also, you might want to look up the Octonions. – user1729 Jul 28 '13 at 19:58
• – Jack Schmidt Jul 28 '13 at 20:15

Such sets are called loops, see

http://en.wikipedia.org/wiki/Loop_(mathematics)

• A loop is not required to have two-sided inverses. – Mikko Korhonen Jul 28 '13 at 18:25
• @m. k.: "which leaves us with closure, identity and invertibility" - I don't see two-sided inverses here! – Boris Novikov Jul 28 '13 at 18:27
• @m. k.: Apropos, there are commutative loops, i.e. having two-sided inverses. – Boris Novikov Jul 28 '13 at 19:39

Hint: Take the integers under subtraction.

(Note: This is a hint, not a solution. You need to alter it slightly to get an identity, as $x-0=x$ but $0-x=-x$.)

• But what is the identity? – Mikko Korhonen Jul 28 '13 at 18:02
• Yet $0 - x = x$ if and only if $x = 0$ – Mikko Korhonen Jul 28 '13 at 18:05
• @m.k. I have removed my last comment and undone my edit - I had forgotten that it was meant to be a hint! – user1729 Jul 28 '13 at 19:41