# Is every finite weak group a group

Definition: Let $W$ be a set and $\circ:W\times W\rightarrow W$ be a function. We say that $(W,\circ)$ is a a weak group iff there exists unique $e\in W$ such that $\forall x\in W[x\circ e=e\circ x=x]$ and for every $x\in W$ there exists a unique element $x^{-1}$ in $W$ such that $x\circ x^{-1}=x^{-1}\circ x=e$. Finally, for every $x,y\in W$, we have: $$x\circ(x^{-1}\circ y)=y\,\,\,\,,(y\circ x^{-1})\circ x=y$$

Question: Is there a finite weak group that is not a group ?

I tried trying playing with Cayley tables for some time to find a finite weak group that is not a group but didn't find any.

Thank you

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@SanathDevalapurkar Yes. This can be inferred from the fact that $\circ$ is a function from $W\times W$ to $W$ –  Amr May 8 '14 at 1:12
Are you claiming $x\circ(x^{-1}\circ y)=y\,\,\,\,,(y\circ x^{-1})\circ x=y$ as part of the axioms? –  Brian Fitzpatrick May 8 '14 at 1:26
@BrianFitzpatrick Yes they are part of the axioms of weak groups –  Amr May 8 '14 at 1:28

Analogous to the finite quaternion group $Q_8$, the set containing the positive and negative generators of the octonions is a finite weak group but not a group.
Every weak group is a group. We already have an identity and inverses, so we need to show that $\circ$ is associative. For all $a,b,c\in W$, we have $b^{-1}\circ (b\circ c)=c$ so that $$(a\circ b)\circ c= (a\circ b) \circ b^{-1} \circ (b\circ c) = a \circ (b\circ c).$$
Edit: I just realized an error. It should be $$(a\circ b)\circ c= (a\circ b) \circ (b^{-1} \circ (b\circ c) )...$$ This does not seem to work anymore.