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Question
The permutation \(P\) is given by
\[P = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&4&5&6&2&1 \end{array}} \right).\]
Determine the order of \(P\), justifying your answer.
Find \({P^2}\).
The permutation group \(G\) is generated by \(P\). Determine the element of \(G\) that is of order 2, giving your answer in cycle notation.
Answer/Explanation
Markscheme
the order is 6 A1
tracking 1 through successive powers of \(P\) returns to 1 after 6 transitions (or equivalent) R1
[2 marks]
\({P^2} = (1{\text{ }}5{\text{ }}4)(2{\text{ }}6{\text{ }}3){\text{ or }}\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 5&6&2&1&4&3 \end{array}} \right)\) (M1)A1
[2 marks]
since \(P\) is of order 6, \({P^3}\) will be of order 2 R1
\({P^3} = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 2&1&4&3&6&5 \end{array}} \right)\) (M1)(A1)
\({P^3} = (1{\text{ }}2)(3{\text{ }}4)(5{\text{ }}6)\) A1
[4 marks]