# Show that there is exactly one binary operation making the set $\{e, x, y\}$ a group with $e$ the identity element

Let $S=\{e, x, y\}$ be a set of three elements. Show that there is exactly one binary operation making the set $S$ a group such that $e$ is the identity element.

So I have produced the multiplication table as follows:

$\begin{array}{c|ccccc} \: & e & x & y \\ \hline e & e & x & y \\ x & x & y & e \\ y & y & e & x \end{array}$

I can see how to explain the first row and column, and also why $xy=e$ and $yx=e$, but I cannot justify $xx=y$ or $yy=x$, other than "because elements cannot be repeated accross rows or columns", which doesn't seem enough to me.

• You said you can justify $xy=e=yx$, so consider what that tells you about $xx$ and $yy$. – hardmath Feb 3 '15 at 19:53

To see why assume that they are repeated. Then you'd have an equation of the form $ab = ac$ or $ab = cb$, which would imply $b = c$ or $a = c$, which cannot happen as you are assuming that $e, x, y$ are distinct.