(i)     Show that the order of \({S_n}\) is \(n!\);

(ii)     List the 6 elements of \({S_3}\) in cycle form;

(iii)     Show that \({S_3}\) is not Abelian;

(iv)     Deduce that \({S_n}\) is not Abelian for \(n \geqslant 3\).

(i)     Write down the matrices M\(_1\), M\(_2\) representing the permutations \((1{\text{ }}2),{\text{ }}(2{\text{ }}3)\), respectively;

(ii)     Find M\(_1\)M\(_2\) and state the permutation represented by this matrix;

(iii)     Find \(\det (\)M\(_1)\), \(\det (\)M\(_2)\) and deduce the value of \(\det (\)M\(_1\)M\(_2)\).

(i)     Use mathematical induction to prove that

\((1{\text{ }}n)(1{\text{ }}n{\text{ }} - 1)(1{\text{ }}n - 2) \ldots (1{\text{ }}2) = (1{\text{ }}2{\text{ }}3 \ldots n){\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}n > 1\).

(ii)     Deduce that every permutation can be written as a product of cycles of length 2.

(i)     1 has \(n\) possible new positions; 2 then has \(n - 1\) possible new positions…

\(n\) has only one possible new position     R1

the number of possible permutations is \(n \times (n - 1) \times  \ldots  \times 2 \times 1\)     R1

\( = n!\)    AG

Note:     Give no credit for simply stating that the number of permutations is \(n!\)

(ii)     \((1)(2)(3);{\text{ }}(1{\text{ }}2)(3);{\text{ }}(1{\text{ }}3)(2);{\text{ }}(2{\text{ }}3)(1);{\text{ }}(1{\text{ }}2{\text{ }}3);{\text{ }}(1{\text{ }}3{\text{ }}2)\)     A2

Notes: A1 for 4 or 5 correct.

If single bracket terms are missing, do not penalize.

Accept \(e\) in place of the identity.

(iii)     attempt to compare \({\pi _1} \circ {\pi _2}\) with \({\pi _2} \circ {\pi _1}\) for two permutations     M1

for example \((1{\text{ }}2)(1{\text{ }}3) = (1{\text{ }}3{\text{ }}2)\)     A1

but \((1{\text{ }}3)(1{\text{ }}2) = (1{\text{ }}2{\text{ }}3)\)     A1

hence \({S_3}\) is not Abelian     AG

(iv)     \({S_3}\) is a subgroup of \({S_n}\),     R1

so \({S_n}\) contains non-commuting elements     R1

\( \Rightarrow {S_n}\) is not Abelian for \(n \geqslant 3\)     AG

[9 marks]

(i)     M\(_1 = \left( {\begin{array}{*{20}{c}} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array}} \right)\), M\(_2 = \left( {\begin{array}{*{20}{c}} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{array}} \right)\)     A1A1

(ii)     M\(_1\)M\(_2 = \left( {\begin{array}{*{20}{c}} 0&0&1&0 \\ 1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{array}} \right)\)     A1

this represents \((1{\text{ }}3{\text{ }}2)\)     A1

(iii)     by, for example, interchanging a pair of rows     (M1)

\(\det (\)M\(_1) = \det (\)M\(_2) =  - 1\)     A1

then \(\det (\)M\(_1\)M\(_2) = ( - 1) \times ( - 1) = 1\)     A1

[7 marks]

(i)     let \({\text{P}}(n)\) be the proposition that

\((1{\text{ }}n)(1{\text{ }}n - 1)(1{\text{ }}n - 2) \ldots (1{\text{ }}2) = (1{\text{ }}2{\text{ }}3 \ldots n){\text{ }}n \in {\mathbb{Z}^ + }\)

the statement that \({\text{P}}(2)\) is true eg \((1{\text{ }}2) = (1{\text{ }}2)\)     A1

assume \({\text{P}}(k)\) is true for some \(k\)     M1

consider \((1{\text{ }}k + 1)(1{\text{ }}k)(1{\text{ }}k - 1)(1{\text{ }}k - 2) \ldots (1{\text{ }}2)\)

\( = (1{\text{ }}k + 1)(1{\text{ }}2{\text{ }}3 \ldots k)\)    M1

then the composite permutation has the following effect on the first \(k + 1\) integers: \(1 \to 2,{\text{ }}2 \to 3 \ldots k - 1 \to k,{\text{ }}k \to 1 \to k + 1,{\text{ }}k + 1 \to 1\)     A1

this is \((1{\text{ }}2{\text{ }}3 \ldots k{\text{ }}k + 1)\)     A1

hence the assertion is true by induction     AG

(ii)     every permutation is a product of cycles     R1

generalizing the result in (i)     R1

every cycle is a product of cycles of length 2     R1

hence every permutation can be written as a product of cycles of length 2     AG

[8 marks]

In part (a)(i), many just wrote down \(n!\) without showing how this arises by a sequential choice process. Part (ii) was usually correctly answered, although some gave their answers in the unwanted 2-dimensional form. Part (iii) was often well answered, though some candidates failed to realise that they need to explicitly evaluate the product of two elements in both orders.

Part (b) was often well answered. A number of candidates found \(2 \times 2\) matrices – this gained no marks.

Nearly all candidates knew how to approach part (c)(i), but failed to be completely convincing. Few candidates seemed to know that every permutation can be written as a product of non-overlapping cycles, as the first step in part (ii).