cayley-tablefinite-groupsgroup-theory
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!
No! Such a Cayley table will not necessarily satisfy the associative law, which is an absolutely critical property in group theory.
You can easily construct a counterexample on 5 elements.
a b c d e b c d e a c a e b d d e b a c e d a c b
Every symbol occurs exactly once in every row or column, and $a$ is the identity element, but this is not associative: $(b*b)*b = a$ while $b*(b*b) = d$. (The multiplication $xy$ is defined by looking up row $x$ and column $y$.)
In general, Cayley tables are not a good way of constructing groups. Even though Cayley tables are usually presented in beginner texts due to their resemblance to the familiar addition/multiplication tables, the associative property is difficult to check in a Cayley table. Instead, groups are usually constructed in terms of some operation which is already known to be associative, such as function composition, integer addition or multiplication, or (once you get off the ground) previously constructed groups.
For any group property, you can always find a Cayley table where there's a duplicate entry and yet that property still holds. Here are examples for each:
Associativity can still hold. $$\begin{array}{c|cc} \times& 0 & 1 \\\hline 0 & 0 & 0\\1 & 0 & 1\end{array}$$
Identity can still hold. (Same example.) $$\begin{array}{c|cc} \times& 0 & 1 \\\hline 0 & 0 & 0\\1 & 0 & 1\end{array}$$
Inverses can still hold. (Here, $a$ and $b$ are inverses of each other.)$$\begin{array}{c|ccc} & e & a & b \\\hline e & e & a & b\\a & a & a & e \\b & b & e & a\end{array}$$
If there's a duplicate row, then $ab=ac$ for some $b\neq c$. Suppose the operator has inverses and associativity. Then we get $a^{-1}ab = a^{-1}ac$ so that $b=c$— contradicting our assumption that $b\neq c$.
So if there's a duplicate row, the operator can either be associative (as shown above), or have inverses (as shown above), but never both.
For confirmation, note that in the example tables above, #1 is associative but not invertible because of 0, and #3 is invertible but not associative because $(bb)a = aa = a$ but $b(ba) = be = b$.)
Groups can't have repeat entries. Therefore, if a table has repeat entries, it's not a group. If it's not a group, then it's not in the green region of this diagram. Visually, you can see that such a table can't be both associative and have inverses at the same time. And you can show that there exist tables with duplicate rows that belong to any other non-green region of this diagram.
Best Answer
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.