Cayley tables for two non-isomorphic groups of order 4.

Solution 1:

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.

Solution 2:

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.