質問: Not sure how to make tables, but: For a binary operation $*$, and set $\{e,a,b,c\}$, in the Cayley table, $a*a$ can be filled with either the identity or an element different from both $e$ and $a$. If in the table the place for $a*a$ is filled with $e$, then the rest of the table can be filled out in two different ways. What are the two different ways? Thanks
回答: Two days ago I computed all Cayley tables for the Gorup $(G,*)$ with $|G|=4$ and came up with 4 distinct Cayley tables (!): For the abelian group of the itegers modulo $4$: $(\mathbb{Z}/(4),+_4)$. $1.$ $$\begin{array}{|c|c|c|c|c|} \hline +_4 & e & a & b & c\\ \hline e & e & a & b & c \\ \hline a & a & b & c & e \\ \hline b & b & c & e & a \\ \hline c & c & e & a & b \\ \hline \end{array}$$ For the Group formed by the $\mathbb{C}$-roots of $z^4=1$: $(\xi_4,\cdot_\mathbb{C})$ $2.$ with $\xi_4=\{ 1,i,-i,-1 \}$ $$\begin{array}{|c|c|c|c|c|} \hline \cdot_\mathbb{C} & e & a & b & c\\ \hline e & e & a & b & c \\ \hline a & a & c & e & b \\ \hline b & b & e & c & a \\ \hline c & c & b & a & e \\ \hline \end{array}$$ $3.$ with $\xi_4=\{1,-1,i,-i \}$ $$\begin{array}{|c|c|c|c|c|} \hline \cdot_\mathbb{C} & e & a & b & c\\ \hline e & e & a & b & c \\ \hline a & a & e & c & b \\ \hline b & b & c & a & e \\ \hline c & c & b & e & a \\ \hline \end{array}$$ And for the Klein-$4$-Group: $(\mathbb{Z}_2\times\mathbb{Z}_2,+_{\mathbb{Z}_2\times\mathbb{Z}_2})$ $4.$ $$\begin{array}{|c|c|c|c|c|} \hline +_{\mathbb{Z}_2\times\mathbb{Z}_2} & e & a & b & c\\ \hline e & e & a & b & c \\ \hline a & a & e & c & b \\ \hline b & b & c & e & a \\ \hline c & c & b & a & e \\ \hline \end{array}$$ The last two are the ones that you are looking for.