There is exactly one group $G$ of four elements, say $G = \{e, a, b, c\}$ satisfying the additional property that $xx = e$ for every $x \in G$. Complete the following group table of $G$.
$$ \begin{array}{c|lcr} & e & a & b & c\\ \hline e & e & a & b & c\\ a & a & & \\ b & b & & \\ c & c & & \end{array}$$
Every $x$ in $G$ satisfies $xx = e,$ so $aa = e, bb = e, cc = e.$
first row: $ab \neq a$ because it's given that $b \neq e.$ Also, $ab \neq b$ since $a \neq e.$ So, $ab = c.$ Thus, $ac = b.$
second row: $ba \neq a$ since $b \neq e.$ So, $ba = c.$ Then, $bc = a.$
third row: $ca \neq a$ since $c \neq e.$ So, $ca = b$ and $cb = a.$
So, $$ \begin{array}{c|lcr} & e & a & b & c\\ \hline e & e & a & b & c\\ a & a & e & c & b\\ b & b & c & e & a\\ c & c & b & a & e \end{array}$$
Please, check my work.
$\endgroup$ 21 Answer
$\begingroup$"Fortunately" there are only two choices for a finite group of order $4$.
One is $(\mathbb{Z}_4,+)$, which is not unipotent.
The second choice, $(\mathbb{Z}_2\times\mathbb{Z}_2,+)$ or Klein four-group, is indeed unipotent and, if you check this Cayley table of it, you'll have your solution confirmed with no even need of relabelling the elements.
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