The unique loop (quasigroup with unit) $L$ of order $5$ satisfying $x^2 = 1$ for all $x \in L$

There are 4 quasi-quaternion groups, strictly $abc, adb, acd, bdc$, (even, odd, even, odd permutations of 3 of $abcd$) with $xy=z, yz=x, zx=y, (xy)z=x(yz)=1$.

They can be visualised on 4 tetrahedral directed graphs, with $x\to y$ implying we follow the arrow, otherwise follow the blank edge.

These can be combined into a tetrahedron with each face given a clockwise or anti-clockwise spin (all faces have the same spin in this case - viewed from outside the tetrahedron). Adding the identity involves adding bidirectional edges and loops.

The idea can be extended to other solids with 3 edges per vertex, e.g. the cube and the dodecahedron.

A Loop with the involution condition you describe has $1$'s along it's main diagonal. Lets remove the $1$ from the Loop to define a related Quasigroup. We maintain the rest of the structure, but need to fill in the now empty entries on the main diagonal. Each row and column are missing a single element and there is only way to satisfy the Latin square condition: the new Quasigroup must be idempotent! Note that we can do the reverse process uniquely too, so there is a one to one correspondence between Loops with this involution condition and Idempotent Quasigroups, so this construction is not so arbitrary.

$\begin{array}{c|ccccc} \ast & 1 & a & b & c & d \\ \hline 1 & 1 & a & b & c & d \\ a & a & 1 & c & d & b \\ b & b & d & 1 & a & c \\ c & c & b & d & 1 & a \\ d & d & c & a & b & 1 \end{array} \Leftrightarrow \begin{array}{c|cccc} ? & a & b & c & d \\ \hline a & & c & d & b \\ b & d & & a & c \\ c & b & d & & a \\ d & c & a & b & \end{array} \Leftrightarrow \begin{array}{c|cccc} \lhd & a & b & c & d \\ \hline a & a & c & d & b \\ b & d & b & a & c \\ c & b & d & c & a \\ d & c & a & b & d \end{array}$

Now that we've done that, the Quasigroup we've constructed has a nice interpretation.

Consider a conjugacy class of elements of order $3$ from the Alternating Group on a set of $4$ elements, $A_4$. Say, $\{(1,2,3),(1,4,2),(1,3,4),(2,4,3)\}$. Define the operation $x\lhd y = xyx^{-1}$ where the implied operations on the right hand side are the usual multiplication of permutations, that is, our operation now conjugates elements.

Let

$a\rightarrow(1,2,3)$

$b\rightarrow(1,4,2)$

$c\rightarrow(2,4,3)$

$d\rightarrow(1,3,4)$

and we get our quasigroup!

For example $a\lhd b=(1,2,3)\lhd(1,4,2) = (1,2,3)(1,4,2)(1,2,3)^{-1}=(1,2,3)(1,4,2)(1,3,2)=(2,4,3)=c$

See also, conjugation and Quandles.