Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
@SanathDevalapurkar Yes. This can be inferred from the fact that $\circ$ is a function from $W\times W$ to $W$ –  Amr May 8 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 at 1:26
    
@BrianFitzpatrick Yes they are part of the axioms of weak groups –  Amr May 8 at 1:28

2 Answers 2

up vote 3 down vote accepted

An example for an infinite weak group which is not a group is the algebra of octonions (http://en.wikipedia.org/wiki/Octonion).

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.