Completing Cayley table for a group

The task is to complete the following Cayley table for a given group. $e$ is of course the identity element. Together with group axioms and the fact that every Cayley table of a group must be a latin square, I arrived at Is it correct? I know there is only one possible table and it sure looks like a latin square but I would appreciate if someone could confirm it's correct or not!

• Think Sudoku! =) – Jose Arnaldo Bebita-Dris Nov 7 '14 at 0:56
• lol yes I did, I checked that it was a Sudoku (or rather a latin square) but does that make it correct? Just wanting to make sure! – user190523 Nov 7 '14 at 1:04
• Being a latin square does not imply itself that it is correct, but your answer is correct. – Derek Holt Nov 7 '14 at 1:14
• Yeah for sure, there exist latin squares which do not represent groups, hence the question. Thanks! – user190523 Nov 7 '14 at 17:22

Completing the Cayley table of order $6$ (if $pq=e$, it must be $qp=e$, etc.), you get a non commutative ($pr\ne rp$) magma, with identity element $e$.
Each element of it has unique inverse (each one is inverse of itself, but $p$ and $q$ mutually inverses).
Finally, if you relabel $e,p,q,r,s,t$ as $1,2,3,4,5,6$, you get the table
$$\begin{array}{c|cccccc} \cdot & 1 & 2 & 3 & 4 & 5 & 6\\ \hline 1 & \color{green}{1} & \color{red}{2} & \color{red}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6}\\ 2 & \color{red}{2} & \color{red}{3} & \color{green}{1} & \color{blue}{5} & \color{blue}{6} & \color{blue}{4}\\ 3 & \color{red}{3} & \color{green}{1} & \color{red}{2} & \color{blue}{6} & \color{blue}{4} & \color{blue}{5}\\ 4 & \color{blue}{4} & \color{blue}{6} & \color{blue}{5} & \color{green}{1} & \color{red}{3} & \color{red}{2}\\ 5 & \color{blue}{5} & \color{blue}{4} & \color{blue}{6} & \color{red}{2} & \color{green}{1} & \color{red}{3}\\ 6 & \color{blue}{6} & \color{blue}{5} & \color{blue}{4} & \color{red}{3} & \color{red}{2} & \color{green}{1} \end{array}$$