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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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  • $\begingroup$ did @achacttn found an answer? $\endgroup$ – scjorge Jul 20 '16 at 1:17
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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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    $\begingroup$ The problem with "counting Cayley tables" is that they mostly only come up at the beginning of one's study of group theory, whereas it takes some group-theoretic sophistication to understand how and what to count, especially when two Cayley tables are to be regarded as distinct. In order to avoid a lot of tables which are permutations of each other, I would have expected the convention that whenever a product of two group elements is an element that hasn't yet appeared above and to the left of it in the table, one should name the new element with the next available letter in the alphabet. $\endgroup$ – Pete L. Clark May 15 '13 at 23:31
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    $\begingroup$ Your second Cayley table violates this by setting $a*a = c$, when we could equally well have put $a*a = b$. The difference, notice, is only that we have interchanged the last two rows and columns of the table. Why do we want to count these as distinct? $\endgroup$ – Pete L. Clark May 15 '13 at 23:32
  • $\begingroup$ I get what you mean, I'm still missing the knowledge to classify by isomorphism these tables. is that where you are going? $\endgroup$ – scjorge May 15 '13 at 23:55
  • $\begingroup$ 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). $\endgroup$ – Jordon Birk Jan 18 '17 at 19:35
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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.

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