All loops of order less than $5$ are groups I am studying loop theory and found that there are no non-trivial loops of order less than $5$. Can anybody help me in this, with some reference of a book or a research article?
A set $Q$ with a binary operation $\circ $ is called a loop if $e \in Q$, where $e$ is the identity element and both the equations $x \circ a = b$ and $a \circ x = b$ has unique solutions, for any $a, b \in Q$. 
By trivial loop I mean it is a group. 

Without loss of generality we denote the loop's elements as $0,1,2,3$ with $0$ the identity and draw the Cayley table:
$$\begin{array}
00&1&2&3\\
1&&&\\
2&&&\\
3&&&\end{array}$$
By the Latin square property of quasigroups (shared with loops) there are only two ways to fill in the two remaining $1$ entries:
$$\begin{array}
00&1&2&3\\
1&&&\\
2&&1&\\
3&&&1\end{array}\qquad\begin{array}
00&1&2&3\\
1&&&\\
2&&&1\\
3&&1&\end{array}$$
The left possibility has only one way to fill in the $2$s and the $3$ by the Latin square property, which in turn forces the $0$s, giving a complete Cayley table:
$$\begin{array}
00&1&2&3\\
1&0&3&2\\
2&3&1&0\\
3&2&0&1\end{array}$$
But this is isomorphic to $\mathbb Z_4$ with $2$ as a generator.
Turning to the other option, we have two ways to place the $2$s while obeying the Latin square property:
$$\begin{array}
00&1&2&3\\
1&&&2\\
2&&&1\\
3&2&1&\end{array}\qquad\begin{array}
00&1&2&3\\
1&2&&\\
2&&&1\\
3&&1&2\end{array}$$
The second option forces the rest of the Cayley table:
$$\begin{array}
00&1&2&3\\
1&2&3&0\\
2&3&0&1\\
3&0&1&2\end{array}$$
which is the familiar presentation of $\mathbb Z_4$, generated by $1$. Once again, we have two possibilities for placing the $3$s in the last unfilled table above, and now both options force the remaining entries:
$$\begin{array}
00&1&2&3\\
1&0&3&2\\
2&3&0&1\\
3&2&1&0\end{array}\qquad\begin{array}
00&1&2&3\\
1&3&0&2\\
2&0&3&1\\
3&2&1&0\end{array}$$
The right table describes $\mathbb Z_4$, generated by $1$, and the left table describes the Klein four-group $\mathbb Z_2^2$. Hence

all loops of order $4$ are groups.

It is easy to check that all loops of order $1,2,3$ are groups by the same method, and I leave this as an exercise.
