(a)     Write down why the table below is a Latin square.

\[\begin{gathered}
  \begin{array}{*{20}{c}}
  {}&d&e&b&a&c
\end{array} \\
  \begin{array}{*{20}{c}}
  d \\
  e \\
  b \\
  a \\
  c
\end{array}\left[ {\begin{array}{*{20}{c}}
  c&d&e&b&a \\
  d&e&b&a&c \\
  a&b&d&c&e \\
  b&a&c&e&d \\
  e&c&a&d&b
\end{array}} \right] \\
\end{gathered} \]

(b)     Use Lagrange’s theorem to show that the table is not a group table.

(a)     Each row and column contains all the elements of the set.     A1A1

[2 marks]

(b)     There are 5 elements therefore any subgroup must be of an order that is a factor of 5     R2

But there is a subgroup \(\begin{gathered}
  \begin{array}{*{20}{c}}
  {}&e&a
\end{array} \\
  \begin{array}{*{20}{c}}
  e \\
  a
\end{array}\left( {\begin{array}{*{20}{c}}
  e&a \\
  a&e
\end{array}} \right) \\
\end{gathered} \) of order 2 so the table is not a group table     R2

Note: Award R0R2 for “a is an element of order 2 which does not divide the order of the group”.

[4 marks]

Total [6 marks]

Part (a) presented no problem but finding the order two subgroups (Lagrange’s theorem was often quoted correctly) was beyond some candidates. Possibly presenting the set in non-alphabetical order was the problem.