(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.