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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

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    did @achacttn found an answer?2016-07-20

2 Answers 2

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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.

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    The simplest way of explaining this is to just start with $a$ in every table, and check multiplying by $a$ however many times until you get a cycle, then check multiplication by $b$, and so on. What you realize by doing this is that, if you just swap out the names (change a to b, b to c, c to a, and so on), then 3 of the 4 tables are actually the same table, in that they just cycle through the pattern,(in different orders, but if you swap the letters, the order doesn't matter).2017-01-18
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If $a * a = e$, then the following are also determined (since each row/column of the Cayley table contains each group element exactly once): $b * a = a * b = c; \quad c * a = a * c = b.$ It follows that the only remaining possibilities for $b*b$ are $e$ and $c$, and we can extend each of these (in exactly one way) to give the Cayley table for a group.