TITLE: The unique loop (quasigroup with unit) $L$ of order $5$ satisfying $x^2 = 1$ for all $x \in L$
QUESTION [10 upvotes]: Recall that a quasigroup is a pair $(Q, \ast)$, where $Q$ is a set and $\ast$ is a binary product $$\ast: Q \times Q \to Q$$ satisfying the Latin square property, namely that for all $x, y \in Q$ there is a unique $a \in Q$ such that $y = ax$ and a unique $b \in Q$ such that $y = x b$, or equivalently, that the multiplication table of $\ast$ is a Latin square. A quasigroup $(Q, \ast)$ is a loop iff it has an identity element, that is an element $1$ such that $1\ast x = x = x\ast 1$ for all $x \in Q$.
This question asks about constructing a loop $(L, \ast)$ on a set $L$ of five elements (denote $L = \{1, a, b, c, d\}$) that satisfies the involution condition $x^2 = 1$ for all $x \in L$. My answer there shows that there is only one such loop up to isomorphism; its multiplication table is:
$$
\begin{array}{c|ccccc}
\ast & 1 & a & b & c & d \\
\hline
   1 & 1 & a & b & c & d \\
   a & a & 1 & c & d & b \\
   b & b & d & 1 & a & c \\
   c & c & b & d & 1 & a \\
   d & d & c & a & b & 1
\end{array}.$$
(Note that it is nonassociative, as $(ab)d = a \neq ac = a(bd)$.)
Even without the involution condition, this example is minimal in the sense that any loop of order $< 5$ is in fact a group. (Up to isomorphism there are six loops of order $5$: This one, the group $(\mathbb{Z}_5, +)$, and four other non-groups.)
So, given that this example is both minimal and unique, it's natural to ask:

Is there a more informative/interesting way to view the loop structure $\ast$ on $L$ than via its multiplication table? That is, does it arise naturally in some other setting?

REPLY [2 votes]: 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.