Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am considering finite magmas $(S,\cdot)$ with $\forall(x,y,z)\in S^3, x\cdot(y\cdot z)=y\cdot(x\cdot z)$. Any finite commutative group is an example of such thing. But in the context (this question on, I am not interested in groups; or at least, not in groups with an efficiently computable inverse.

I am wondering if/how the classical magma-to-group classification simplifies for such structures. magma to group

share|cite|improve this question
Actually, existence of identity is enough for it to imply commutativity. – tomasz Jul 8 '12 at 12:48

About quasigroups

The submagma $M_S$ of $S\to S$ generated by left multiplication $\phi_x: y\mapsto x\cdot y$ is commutative. If $S$ is a finite quasigroup, $M_S$ is only composed of permutations (as $\phi_x(x\backslash y)=y$ for all $y$), so it is a (commutative) subgroup of the symmetric group $\mathcal S_n$. In particular for each $x$ there is an $x^{-1}$ such that $x^{-1}\cdot(x\cdot y)=y$.

Every commutative subgroup $H$ of $\mathcal S_n$ ($|H|\ge 3$) can be associated to a quasigroup without identity in this way: pick a permutation $\pi$ of the $|H|$ elements of $H$ distinct from the identity such that $\pi(1)=1$. Then the operation $x\cdot y=\pi^{-1}(x\circ\pi(y))$ defines a quasigroup structure, with $x\backslash y=x^{-1}\cdot y$ and $x/y=\pi(x)\circ\pi(y)^{-1}$. Obviously $1\cdot x=x\ne x\cdot 1$.

In fact this is exactly the set of quasigroups $Q$ without identity such that $x\mapsto \phi_x$ is injective, since necessarily $Q=M_Q$ and $\pi(x)=x\cdot 1$.

If $H$ has an element of order greater than 2, we can choose $\pi(x)=x^{-1}$ and $\cdot$ boils down to left division in $H$: $x\cdot y=x^{-1}\circ y$. So $C_3$ equipped with left division is a simple example of a quasigroup satisfying the condition.

share|cite|improve this answer

It is related to commutativity: if there's an identity element, then it implies it (by putting $z=e$), if we have associativity, then it is implied by it (obviously). On the other hand, the property along with commutativity implies associativity (it is easy to check), so a loop with this property is already a commutative group, and any magma with identity and this property is automatically a commutative monoid.
enter image description here

Any finite set with operation of right projection $a\cdot b=b$ has the property and is associative, but not divisible and does not have identity (if the set has at least 2 elements), so that takes care of two arrows. (It is also not commutative.)

A finite linear order with the operation $a\vee b=\max(a,b)$ is a noninvertible monoid with the property, which takes care of the lower right one.

As per Generic Human's answer, a simple example of a quasigroup without identity (albeit with a left identity) that is nonassociative, but having the property in question, is the cyclic group $\mathbf Z_3$ with the operation $x\cdot y=y-x$ (or any commutative group with an element of order at least 3 with the same operation), so that is the complete picture.

I'm opening a community wiki, so others can contribute.

share|cite|improve this answer

Your Answer


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.