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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.

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  • $\begingroup$ You said you can justify $xy=e=yx$, so consider what that tells you about $xx$ and $yy$. $\endgroup$ – hardmath Feb 3 '15 at 19:53
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Saying that elements cannot be repeated across rows and columns is perfectly valid as long as you explain why they cannot be repeated.

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.

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