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The task is to complete the following Cayley table

cayley

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

cay

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!

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  • $\begingroup$ Think Sudoku! =) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 7 '14 at 0:56
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    $\begingroup$ 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! $\endgroup$ – user190523 Nov 7 '14 at 1:04
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    $\begingroup$ Being a latin square does not imply itself that it is correct, but your answer is correct. $\endgroup$ – Derek Holt Nov 7 '14 at 1:14
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    $\begingroup$ Yeah for sure, there exist latin squares which do not represent groups, hence the question. Thanks! $\endgroup$ – user190523 Nov 7 '14 at 17:22
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I agree with your answer.

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

$$\begin{array}{c|cccccc} \cdot & 1 & 2 & 3 & 4 & 5 & 6\\ \hline 1 & \color{green}{1} & \color{red}{2} & \color{red}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6}\\ 2 & \color{red}{2} & \color{red}{3} & \color{green}{1} & \color{blue}{5} & \color{blue}{6} & \color{blue}{4}\\ 3 & \color{red}{3} & \color{green}{1} & \color{red}{2} & \color{blue}{6} & \color{blue}{4} & \color{blue}{5}\\ 4 & \color{blue}{4} & \color{blue}{6} & \color{blue}{5} & \color{green}{1} & \color{red}{3} & \color{red}{2}\\ 5 & \color{blue}{5} & \color{blue}{4} & \color{blue}{6} & \color{red}{2} & \color{green}{1} & \color{red}{3}\\ 6 & \color{blue}{6} & \color{blue}{5} & \color{blue}{4} & \color{red}{3} & \color{red}{2} & \color{green}{1} \end{array}$$

that you can check to be associative by using this brute-force Matlab script.

On the other hand, the above is a quasigroup since the Cayley table is a latin square and an associative quasigroup has identity element too.

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