# 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.${}$ Commented Jul 28, 2013 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. Commented Jul 28, 2013 at 18:17
• Just make the product of any two nonidentity elements equal the identity. Commented Jul 28, 2013 at 19:58
• Also, you might want to look up the Octonions. Commented Jul 28, 2013 at 19:58
• Commented Jul 28, 2013 at 20:15

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? Commented Jul 28, 2013 at 18:02
• Yet $0 - x = x$ if and only if $x = 0$ Commented Jul 28, 2013 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! Commented Jul 28, 2013 at 19:41

Such sets are called loops, see

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

• A loop is not required to have two-sided inverses. Commented Jul 28, 2013 at 18:25
• @m. k.: "which leaves us with closure, identity and invertibility" - I don't see two-sided inverses here! Commented Jul 28, 2013 at 18:27
• @m. k.: Apropos, there are commutative loops, i.e. having two-sided inverses. Commented Jul 28, 2013 at 19:39