0

If $G$ has inverse of all elements, then $G$ is a group. $($ true / false$) ?$

I know a group satisfies closure property, associativity. It has a unique identity element and inverses for all elements.

I can't find any counter example to prove the statement false. If the statement is true, then how existence of inverses of all elements imply that $G$ is a group?

Mathaddict
  • 2,300
  • 1
    You haven't given enough information to define $G$. Is $G$ a set equipped with a binary operation? Is $G$ closed under the binary operation? Anything else? – parsiad Jan 12 '19 at 19:38
  • 1
    The unit sphere in $\mathbb R^8$ has a product operation with inverses that is not associative. – Matt Samuel Jan 12 '19 at 19:38
  • @MattSamuel Is there a (smallish) finite substructure of $\mathbb{R}^8$ that also serves as a counterexample? – Mike Pierce Jan 12 '19 at 19:39
  • @Mike Probably. Rather than trying to find that though it would probably be easier to directly construct a finite example. – Matt Samuel Jan 12 '19 at 19:40
  • 1
    The question is a bit vague in terms of restriction on set $G$. For example, we can have a trivial example as $G=\Bbb{Z}-{0}$, then we can say $G$ has additive inverses but is not closed, hence not a group. – Anurag A Jan 12 '19 at 19:42
  • @Mike For example, let $G={e, a, a^{-1}, b, b^{-1}}$. $e$ is the identity. Besides the inverse pairs implied by the notation, you can define products arbitrarily. Should be easy to make it nonassociative. – Matt Samuel Jan 12 '19 at 19:45

1 Answers1

0

No, groupoids (https://en.wikipedia.org/wiki/Groupoid?wprov=sfla1) are a counterexample.

ecrin
  • 1,423