0
$\begingroup$

My question is generated from the following question:

It turns out that the inverse of product with an assumption of inverse existence is a necessary condition of associative. Then is there any set with a binary composition rule satisfies $(ab)^{-1}=b^{-1}a^{-1}\,$ but fails associate? The answer is yes, see the following Cayley table:

$$ \begin{array}{c|lcr} & 1 & 2 & 3 \\ \hline 1 & 1 & 2 & 3 \\ 2 & 2 & 1 & 2 \\ 3 & 3 & 2 & 1 \end{array} $$ The above set has unique inverse and unique identity but has overlapping number in the same column or row. Then I think what if the Cayley table keeps all row and column entries being different which also has a unique identity and unique inverse, can it be a group? The answer is no, see the following example: $$ \begin{array}{c|lcr} & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 \\ 2 & 2 & 1 & 4 & 5 & 3 \\ 3 & 3 & 5 & 1 & 2 & 4 \\ 4 & 4 & 3 & 5 & 1 & 2 \\ 5 & 5 & 4 & 2 & 3 & 1 \end{array} $$ which is not commute and has order 5. This indicates the set cannot be a group since group with order 5 only can be isomorphic to $\mathbb{Z}_5$. However, this Cayley table fails the inverse of product property, e.g. $(2\cdot 3)^{-1}=5^{-1}=5$ while $3^{-1}2^{-1}=3\cdot 2=4$.

Then I am thinking can a set with a binary composition law, a unique identity, unique inverse property, Cayley table having no overlapping element in any row and column and the inverse of product be a group?

Concisely, Cayley table having no overlapping element indicates $\forall a\in S,\,(a\cdot\bullet):S\to S$, and $(\bullet\cdot a):S\to S$ are two bijection. Therefore, the set is a Quasi-group, together with a unit which is called a loop.

Then the question is reduced to ask whether a loop with inverse property and inverse product rule, i.e. $(ab)^{-1}=b^{-1}a^{-1}$ form a group?

Could you find a counterexample? Or just prove it.

On the other hand, what can be an equivalent assumption of the associativity rule.

Thanks.

$\endgroup$
1
$\begingroup$

Any Moufang loop has this property.

Examples of Moufang loops are the nonzero octonions under multiplication or subloops of that. The smallest nonassociative Moufang loop has size 12.

An introduction on Moufang loops can be found for example here.

$\endgroup$
  • $\begingroup$ Yes, you are correct. For all Moufang loops, they all have the inverse product properties but may not have the associativity. $\endgroup$ – Hamio Jiang Jan 28 '18 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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