What are the commutative quasigroups satisfying $a/b=b/a$?
There's a harder question lurking behind this question that was just asked. The context is quasigroup theory. A commutative quasigroup can be defined as a set $Q$ together with commutative binary operation $*$ such that for all $a,b \in Q$, there is a unique "solution" $s \in Q$ solving $s*a=b$. We write $b/a$ for the unique such $s$. The linked question (essentially) asks if there exists a commutative quasigroup satisfying the identity $a/b=b/a$. (Yes, for example $\mathbb{Z}/2\mathbb{Z}$ has this property with respect to addition.) What I'd like to know is, can we usefully characterize all commutative quasigroups satisfying this identity, including the non-associative ones?
Ideas, anyone?
Solution 1:
Updating my previously (wrong!) response.
Let $(Q,\cdot)$ be a commutative quasigroup. TFAE:
- For all $x,y \in Q$ $xy=x/y$.
- For all $x,y \in Q$ $x/y=y/x$.
First note, $(Q,\cdot,\backslash, /)$ is a quasigroup if the following are satisfied for all $x,y \in Q$ $$ x(x\backslash y) = y = x\backslash (xy),\\ (y/x)x = y = (yx)/x. $$ This is an equivalent definition (to unique solutions of $ax=b$ and $ya=b$), but now we are insured that quasigroups form a variety.
Now, it is straightforward to show that if $Q$ is commutative, then $x/y=y\backslash x$. Using this, it is now easy to see that $x/(x/y)=y$ which implies $x*y=x/y$. The other implication is immediate. Again, both directions rely on commutativity.
If you include an identity in your assumptions, ($i.e.$ Q is a loop), then you have that $x=x^{-1}$. So your loop is power associative (it is not necessarily diassociative!). If you add associativity (as already stated), Q is an elementary abelian $2$-group.