Non-associative version of a group satisfying these identities: $(xy)y^{-1}=y^{-1}(yx)=x$

Solution 1:

If for all $x,y \in G$, one has $(xy)y^{-1}=y^{-1}(yx)=x$, then $G$ is a special type of quasi-group that I'll call a SIP-quasigroup (“symmetric inverse property quasigroup”). However, there are quasi-groups without such inverse functions; I give an infinite family of examples due to Belousov that do have an inverse property (but not a symmetric one).

If you assume an identity, then you get exactly the IP-loops (“inverse property loops”) which are reasonably well studied.

All SIP-quasigroups are quasigroups

Lemma: ${(x^{-1})}^{-1}=x$ and $(xy^{-1})y = y(y^{-1}x) = x$

Proof: [by user1] $$\begin{array}{rl} (x^{-1})^{-1} &=\left((x^{-1})^{-1}x\right)x^{-1} \\ &=\left((x^{-1})^{-1}‌​\left(x^{-1}(xx)\right)\right)x^{-1} \\ &=(xx)x^{-1} \\ &=x \end{array}$$ Then substituting $y^{-1}$ instead of $y$ into the defining axioms one gets $x = (xy^{-1})(y^{-1})^{-1} = (xy^{-1}) y$ and similarly on the other side. $\square$

Proposition: $G$ is a quasi-group.

Proof: [by Berci] We need to show $ax=b$ and $ya=b$ have unique solutions $x,y \in G$. However, the defining axioms of a SIP-quasigroup give that $x=a^{-1}(ax) = a^{-1}(b)$ and $y=(ya)a^{-1} = (b)a^{-1}$, so those are the only possible solutions. Since $a(a^{-1}b) = b$ and $(ba^{-1})a=b$ by the lemma, these are in fact solutions, and $G$ is a quasi-group. $\square$

A special type of quasigroup are the IP-quasigroup (“inverse property quasigroups”). These are the quasigroups $G$ with functions $\lambda:G \to G$ and $\rho:G \to G$ such that $x^\lambda (xy) = (yx) {}^\rho x = y$. A loop with both those properties is called an IP-loop, and every IP-loop satisfies $x^\lambda = {}^\rho x$.

Proposition: Every SIP-quasigroup is an IP-quasigroup.

Proof: Simply take $x^\lambda = {}^\rho x = x^{-1}$.

But not all quasigroups are SIP-quasigroups

We give a family of examples of IP-quasigroups that are not SIP-quasigroups.

Let $A$ be an abelian group that has an element of order not 1 or 2, and define a multiplication on $G=A \times A$ by $(a_1,b_1) \cdot (a_2,b_2) = (a_1 + a_2, b_2-b_1)$. $G$ is a quasi-group (but not a loop) with both left and right inverse properties: define $x^\lambda = (-a,-b)$ and ${}^\rho x =(-a,b)$ so that $x^\lambda (xy) = (yx) {}^\rho x = y$.

Verifications:

• $G$ is a quasi-group.

Proof: Let $a,b,c,d \in A$ and we try to uniquely solve $(a,b) \cdot(x,y) = (c,d)$. The left hand side is $(a+x,y-b)$ so the unique solution is $x=c-a$ and $y=b+d$. Hence $G$ is a quasi-group. $\square$

• $G$ is not a loop.

Proof: Suppose by way of contradiction that $(x,y)$ is an identity. Then $(a,b) \cdot (x,y) = (a,b)$ implies $x=0$, $y=2b$, but this depends on $b$, so $(x,y)$ is not an identity. $\square$

• $G$ satisfies the left and right inverse properties, so is an IP-quasigroup.

Proof: These are just calculations: $$(-a,-b) \cdot ( (a,b) \cdot (c,d) ) = (-a,-b) \cdot (a+c,d-b) = (-a+a+c,d-b-(-b)) = (c,d)$$ and $$((c,d) \cdot (a,b)) \cdot(-a,b) = (c+a,b-d) \cdot (-a,b) = (c+a-a,b-(b-d)) =(c,d). \square$$

• $G$ does not satisfy the original questions axioms for any choice of $a^{-1}$.

Proof: The left and right inverses were uniquely determined. Set $(a,b)^{-1} = (x,y)$. Then $(a,b)^{-1} ( (a,b) \cdot (c,d) ) = (x,y) \cdot (a+c,d-b) = (x+a+c,d-b-y)$ so that $x=-a$ and $y=-b$. However, $((c,d) \cdot(a,b))\cdot(x,y) = (c+a,b-d) \cdot (x,y) = (c+a+x,y-(b-d))$ so that $x=-a$ and $y=b$. Unless $b=-b$, these differ. $\square$

• $G$ is an IP-quasigroup that is not a SIP-quasigroup.

Proof: Since the left and right inverses are uniquely defined, the only way for $x^{-1}$ to exist is if the two inverse agree, but they disagree on any element $(a,b)$ with $b$ not of order 1 or 2. $\square$

A loop is a SIP-quasigroup iff IP-loop

If we assume the existence of an identity, then the SIP axioms become equivalent to the IP-loop axioms.

Proposition: For loops $G$, the following are equivalent: * $G$ is a SIP-quasigroup * $G$ is an IP-quasigroup * $G$ is an IP-loop

Proof: This is simply because in a loop $x^\lambda = {}^\rho x$ so that inverses are automatically symmetric. $\square$

SIP-quasigroup that is not a loop

While the axioms for a SIP-quasigroup with identity are equivalent to the standard notion of "IP-loop" or "loop with both left and right inverse properties", a SIP-quasigroup need not have an identity. For example:

$$\begin{array}{c|ccc} \times & 1 & 2 & 3 \\ \hline 1 & 1 & 3 & 2 \\ 2 & 3 & 2 & 1 \\ 3 & 2 & 1 & 3 \end{array}$$

In this quasi-group we have $x^2 = x$ and $xy=yx$ and $x^{-1}=x$ and $xy=z$ as long as $x,y,z$ are distinct (a so-called Steiner quasigroup). Hence it satisfies the original post's axioms, but has no identity.

Frequency of SIP-quasigroups

The multiplication table of a quasigroup is “Latin square” and if we sample uniformly at random from Latin squares we get one (amongst many) notion of frequency of quasigroups. Here are the observed frequencies after 10000 samples at each order:

$\begin{array}{c|rrrrrrrrrr} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline Group & 100\% & 100\% & 25\% & 2\% & & & & & & \\ SIP-quasigroup & & & 25\% & 4\% & & & & & & \\ IP-quasigroup & & & 50\% & 8\% & & & & & & \\ RIP-quasigroup & & & & 44\% & 10\% & & & & & \\ quasigroup & & & & 42\% & 90\% & 100\% & 100\% & 100\% & 100\% & 100\% \\ \end{array}$

A quasigroup is counted only in the first line it qualifies, so that “SIP-quasigroup” here is short for “SIP-quasigroup that is not a group”. “RIP-quasigroup” normally stands for quasigroup with right inverse property, but I use it as shorthand for “quasigroup with right inverse property or left inverse property, but not both.” I don't report percentages less than 0.5%. In particular, loops that aren't groups never broke the 0.5% barrier.

Example quasigroups

Wikipedia has several examples of quasigroups, and it might be nice to know which ones are SIP-quasigroups. I consider groups under division, Steiner quasigroups, and Moufang loops.

Proposition: A group under division is a RIP-quasigroup. It is an IP-quasigroup iff it is a SIP-quasigroup iff the group is abelian. It is a loop iff it is an IP-loop iff the group is an elementary abelian 2-group.

Proof: Define $g\ast h = g h^{-1}$ then $a \ast x = b$ has unique solution $x=b^{-1}a$ and $y \ast a = b$ has unique solution $y=ba$, so $(G,\ast)$ is a quasigroup. Suppose that $x$ is an identity, then $g = x\ast g = xg^{-1}$ implies $x=g^2$ must be constant (and so equal to the identity of the group). If $x \ast (g \ast h) = h$, then solving for $x$ one gets $x=hgh^{-1}$ so that $(G,\ast)$ is a LIP-quasigroup iff $G$ is abelian, and then $x^\lambda = x$. If $(h \ast g ) \ast y = h$, then solving for $y$ one gets $y=g$ so that ${}^\rho g =g$ is always a right inverse. $\square$

Proposition: A Steiner quasigroup is a commutative, idempotent SIP-quasigroup that is not a loop.

Proof: If $x \neq y$, then there is a unique Steiner triple $\{x,y,z\}$ and $xy=yx = z$, so we get that $x(xy) = (yx)x = y$ since $xz=zx=y$. If $x=y$ then $x(xx) = xx = x$ since Steiner quasigroups are idempotent. Note that for any $x$, $xx=x$, but that there is a $y$ with $xy=z$ and $x,y,z$ distinct, so $x$ cannot be the identity. $\square$

Proposition: A Moufang loop is an IP-loop, so also a SIP-quasigroup.

Indeed Moufang loops are exactly the loops that remain IP-loops even after scrambling the rows, columns, and labels.