Show Latin Square is not a group.

Question

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?

Answer 1

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.

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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.

Answer 2

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 $u=v$.

Answer 3

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.