The permutation ${p_1}$ of the set {1, 2, 3, 4} is defined by

$${p_1} = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   2&4&1&3 \end{array}} \right)$$

(a)     (i)     State the inverse of ${p_1}$.

  (ii)     Find the order of ${p_1}$.

(b)     Another permutation ${p_2}$ is defined by

$${p_2} = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   3&2&4&1 \end{array}} \right)$$

  (i)     Determine whether or not the composition of ${p_1}$ and ${p_2}$ is commutative.

  (ii)     Find the permutation ${p_3}$ which satisfies

$${p_1}{p_3}{p_2} = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   1&2&3&4 \end{array}} \right){\text{.}}$$

(a)     (i)     the inverse is

$\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   3&1&4&2 \end{array}} \right)$     A1

(ii)     EITHER

$1 \to 2 \to 4 \to 3 \to 1$ (is a cycle of length 4)     R3

so ${p_1}$ is of order 4     A1     N2

OR

consider

$p_1^2 = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   4&3&1&2 \end{array}} \right)$     M1A1

it is now clear that

$p_1^4 = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   1&2&3&4 \end{array}} \right)$     A1

so ${p_1}$ is of order 4     A1     N2

[5 marks]

(b)     (i)     consider

${p_1}{p_2} = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   2&4&1&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   3&2&4&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   1&4&3&2 \end{array}} \right)$     M1A1

${p_2}{p_1} = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   3&2&4&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   2&4&1&3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   2&1&3&4 \end{array}} \right)$     A1

composition is not commutative     A1

Note: In this part do not penalize candidates who incorrectly reverse the order both times.

(ii)     EITHER

pre and postmultiply by $p_1^{ - 1}$, $p_2^{ - 1}$to give

${p_3} = p_1^{ - 1}p_2^{ - 1}$     (M1)(A1)

$= \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   3&1&4&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   4&2&1&3 \end{array}} \right)$     A1

$= \left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   2&1&3&4 \end{array}} \right)$     A1

OR

starting from

$\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   2&4&1&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   {}&{}&{}&{} \end{array}} \right)\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   3&2&4&1 \end{array}} \right)$     M1

successively deducing each missing number, to get

$\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   2&4&1&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   2&1&3&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}}   1&2&3&4 \\   3&2&4&1 \end{array}} \right)$     A3

[8 marks]

Total [13 marks]

Many candidates scored well on this question although some gave the impression of not having studied this topic. The most common error in (b) was to believe incorrectly that ${p_1}{p_2}$ means ${p_1}$ followed by ${p_2}$. This was condoned in (i) but penalised in (ii). The Guide makes it quite clear that this is the notation to be used.