Show Latin Square is not a group. 

If we fix the first two rows in the above figure, then there are many ways to fill in the remaining rows to obtain a Latin square. Show that none of these Latin squares is the multiplication table of a group.
Hint: Index the first two rows by $x$ and $y$, the first two columns by $u$ and $v$, all under the assumption that we have the multiplication table of a group of order $5$. Show that $(v^{-1}u)^2=1$.

I am not sure how to start this problem. How would we use the hint if we don't know what $u$ and $v$ actually are?
 A: You are given that $xu=a=yv$ as well as $yu=b=xv$. The first equation $xu=yv$ implies
$$
vu^{-1}=y^{-1}x,
$$
but the second equation $yu=xv$ gives
$$
vu^{-1}=x^{-1}y.
$$
Multiplying these two together gives
$$
(vu^{-1})^2=y^{-1}xx^{-1}y=1.
$$
In a group of order five this implies $vu^{-1}=1$, and also $v=u$, which is absurd.

The tell tale sign that something is wrong is to look at the permutation that takes the first row to the second. We see that the permutation has cycle decomposition $(ab)(ced)$. This cannot happen in a group table, because we get the second row from the first by multiplying it by $yx^{-1}$. That element has some order $\ell$, and consequently we should only see cycles of length $\ell$ instead of a mixture of a 2-cycle and a 3-cycle.
A: We have $xu=a$, $yu=b$, $xv=b$, $yv=a$ and thus $u=x^{-1}a$, $u=y^{-1}b$, $v^{-1}=b^{-1}x$, $v^{-1}=a^{-1}y$. Thus $\left(v^{-1}u\right)^2=v^{-1}uv^{-1}u=\left(b^{-1}x\right)\left(x^{-1}a\right)\left(a^{-1}y\right)\left(y^{-1}b\right)=b^{-1}\left(xx^{-1}\right)\left(aa^{-1}\right)\left(yy^{-1}\right)b=1.$ A group of odd order $5$ has no element of order $2$; hence $v^{-1}u=1$, and thus $u=v$.
A: If this occurs in a group multiplication table
$$
\begin{array}{c|cc}
& u & v \\
\hline
x & a & b \\
y & b & a \\
\end{array}
$$
then, by left multiplying by $x^{-1}$ and right multiplying by $u^{-1}$, we find this also occurs in the group's multiplication table
$$
\begin{array}{c|cc}
& uu^{-1} & vu^{-1} \\
\hline
x^{-1}x & x^{-1}au^{-1} & x^{-1}bu^{-1} \\
x^{-1}y & x^{-1}bu^{-1} & x^{-1}au^{-1} \\
\end{array}
\xrightarrow{\text{which simplifies to}}
\begin{array}{c|cc}
& \mathrm{id} & g \\
\hline
\mathrm{id} & \mathrm{id} & g \\
g & g & \mathrm{id} \\
\end{array}
$$
which is a subgroup of order $2$.  If this is in a group of order $5$, we violate Lagrange's Theorem.
This generalizes to subgroups of any size: (a) group tables always decompose into translates (double cosets) of their subgroups, and (b) subsquares are always translates of subgroups.
