# IB DP Maths Topic 8.10 Permutations under composition of permutations HL Paper 3

## Question

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

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

## Markscheme

(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]

## Examiners report

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.

## Question

Consider the set $$A$$ consisting of all the permutations of the integers $$1,2,3,4,5$$.

Two members of $$A$$ are given by $$p = (1{\text{ }}2{\text{ }}5)$$ and $$q = (1{\text{ }}3)(2{\text{ }}5)$$.

Find the single permutation which is equivalent to $$q \circ p$$.

[4]
a.

State a permutation belonging to $$A$$ of order

(i)     $$4$$;

(ii)     $$6$$.

[3]
b.

Let $$P =$$ {all permutations in $$A$$ where exactly two integers change position},

and $$Q =$$ {all permutations in $$A$$ where the integer $$1$$ changes position}.

(i)     List all the elements in $$P \cap Q$$.

(ii)     Find $$n(P \cap Q’)$$.

[4]
c.

## Markscheme

$$q \circ p = (1{\text{ }}3)(2{\text{ }}5)(1{\text{ }}2{\text{ }}5)$$     (M1)

$$= (1{\text{ }}5{\text{ }}3)$$     M1A1A1

Note:     M1 for an answer consisting of disjoint cycles, A1 for $$(1{\text{ }}5{\text{ }}3)$$,

A1 for either $$(2)$$ or $$(2)$$ omitted.

Note:     Allow $$\left( {\begin{array}{*{20}{c}} 1&2&3&4&5 \\ 5&2&1&4&3 \end{array}} \right)$$

If done in the wrong order and obtained $$(1{\text{ }}3{\text{ }}2)$$, award A2.

[4 marks]

a.

(i)     any cycle with length $$4$$ eg ($$1234$$)     A1

(ii)     any permutation with $$2$$ disjoint cycles one of length $$2$$ and one of length $$3$$ eg ($$1$$ $$2$$) ($$3$$ $$4$$ $$5$$)     M1A1

Note:     Award M1A0 for any permutation with $$2$$ non-disjoint cycles one of length $$2$$ and one of length $$3$$.

Accept non cycle notation.

[3 marks]

b.

(i)     ($$1$$, $$2$$), ($$1$$, $$3$$), ($$1$$, $$4$$), ($$1$$, $$5$$)     M1A1

(ii)     ($$2$$ $$3$$), ($$2$$ $$4$$), ($$2$$ $$5$$), ($$3$$ $$4$$), ($$3$$ $$5$$), ($$4$$ $$5$$)     (M1)

6     A1

Note:     Award M1 for at least one correct cycle.

[4 marks]

Total [11 marks]

c.

## Examiners report

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

a.

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

b.

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

c.

## Question

Consider the set $$A$$ consisting of all the permutations of the integers $$1,2,3,4,5$$.

Two members of $$A$$ are given by $$p = (1{\text{ }}2{\text{ }}5)$$ and $$q = (1{\text{ }}3)(2{\text{ }}5)$$.

Find the single permutation which is equivalent to $$q \circ p$$.

[4]
a.

State a permutation belonging to $$A$$ of order

(i)     $$4$$;

(ii)     $$6$$.

[3]
b.

Let $$P =$$ {all permutations in $$A$$ where exactly two integers change position},

and $$Q =$$ {all permutations in $$A$$ where the integer $$1$$ changes position}.

(i)     List all the elements in $$P \cap Q$$.

(ii)     Find $$n(P \cap Q’)$$.

[4]
c.

## Markscheme

$$q \circ p = (1{\text{ }}3)(2{\text{ }}5)(1{\text{ }}2{\text{ }}5)$$     (M1)

$$= (1{\text{ }}5{\text{ }}3)$$     M1A1A1

Note:     M1 for an answer consisting of disjoint cycles, A1 for $$(1{\text{ }}5{\text{ }}3)$$,

A1 for either $$(2)$$ or $$(2)$$ omitted.

Note:     Allow $$\left( {\begin{array}{*{20}{c}} 1&2&3&4&5 \\ 5&2&1&4&3 \end{array}} \right)$$

If done in the wrong order and obtained $$(1{\text{ }}3{\text{ }}2)$$, award A2.

[4 marks]

a.

(i)     any cycle with length $$4$$ eg ($$1234$$)     A1

(ii)     any permutation with $$2$$ disjoint cycles one of length $$2$$ and one of length $$3$$ eg ($$1$$ $$2$$) ($$3$$ $$4$$ $$5$$)     M1A1

Note:     Award M1A0 for any permutation with $$2$$ non-disjoint cycles one of length $$2$$ and one of length $$3$$.

Accept non cycle notation.

[3 marks]

b.

(i)     ($$1$$, $$2$$), ($$1$$, $$3$$), ($$1$$, $$4$$), ($$1$$, $$5$$)     M1A1

(ii)     ($$2$$ $$3$$), ($$2$$ $$4$$), ($$2$$ $$5$$), ($$3$$ $$4$$), ($$3$$ $$5$$), ($$4$$ $$5$$)     (M1)

6     A1

Note:     Award M1 for at least one correct cycle.

[4 marks]

Total [11 marks]

c.

## Examiners report

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

a.

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

b.

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

c.

## Question

The set of all permutations of the elements $$1,{\text{ }}2,{\text{ }} \ldots 10$$ is denoted by $$H$$ and the binary operation $$\circ$$ represents the composition of permutations.

The permutation $$p = (1{\text{ }}2{\text{ }}3{\text{ }}4{\text{ }}5{\text{ }}6)(7{\text{ }}8{\text{ }}9{\text{ }}10)$$ generates the subgroup $$\{ G,{\text{ }} \circ \}$$ of the group $$\{ H,{\text{ }} \circ \}$$.

Find the order of $$\{ G,{\text{ }} \circ \}$$.

[2]
a.

State the identity element in $$\{ G,{\text{ }} \circ \}$$.

[1]
b.

Find

(i)     $$p \circ p$$;

(ii)     the inverse of $$p \circ p$$.

[4]
c.

(i)     Find the maximum possible order of an element in $$\{ H,{\text{ }} \circ \}$$.

(ii)     Give an example of an element with this order.

[3]
d.

## Markscheme

the order of $$(G,{\text{ }} \circ )$$ is $${\text{lcm}}(6,{\text{ }}4)$$     (M1)

$$= 12$$     A1

[2 marks]

a.

$$\left( 1 \right){\rm{ }}\left( 2 \right){\rm{ }}\left( 3 \right){\rm{ }}\left( 4 \right){\rm{ }}\left( 5 \right){\rm{ }}\left( 6 \right){\rm{ }}\left( 7 \right){\rm{ }}\left( 8 \right){\rm{ }}\left( 9 \right){\rm{ }}\left( {10} \right)$$     A1

Note:     Accept ( ) or a word description.

[1 mark]

b.

(i)     $$p \circ p = (1{\text{ }}3{\text{ }}5)(2{\text{ }}4{\text{ }}6)(7{\text{ }}9)(810)$$     (M1)A1

(ii)     its inverse $$= (1{\text{ }}5{\text{ }}3)(2{\text{ }}6{\text{ }}4)(7{\text{ }}9)(810)$$     A1A1

Note:     Award A1 for cycles of 2, A1 for cycles of 3.

[4 marks]

c.

(i)     considering LCM of length of cycles with length $$2$$, $$3$$ and $$5$$     (M1)

$$30$$     A1

(ii)     eg$$\;\;\;(1{\text{ }}2)(3{\text{ }}4{\text{ }}5)(6{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}10)$$     A1

Note:     allow FT as long as the length of cycles adds to $$10$$ and their LCM is consistent with answer to part (i).

Note: Accept alternative notation for each part

[3 marks]

Total [10 marks]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

## Question

Let $$\{ G,{\text{ }} \circ \}$$ be the group of all permutations of $$1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6$$ under the operation of composition of permutations.

Consider the following Venn diagram, where $$A = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4\} ,{\text{ }}B = \{ 3,{\text{ }}4,{\text{ }}5,{\text{ }}6\}$$.

(i)     Write the permutation $$\alpha = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&4&6&2&1&5 \end{array}} \right)$$ as a composition of disjoint cycles.

(ii)     State the order of $$\alpha$$.

[3]
a.

(i)     Write the permutation $$\beta = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 6&4&3&5&1&2 \end{array}} \right)$$ as a composition of disjoint cycles.

(ii)     State the order of $$\beta$$.

[2]
b.

Write the permutation $$\alpha \circ \beta$$ as a composition of disjoint cycles.

[2]
c.

Write the permutation $$\beta \circ \alpha$$ as a composition of disjoint cycles.

[2]
d.

State the order of $$\{ G,{\text{ }} \circ \}$$.

[2]
e.

Find the number of permutations in $$\{ G,{\text{ }} \circ \}$$ which will result in $$A$$, $$B$$ and $$A \cap B$$ remaining unchanged.

[2]
f.

## Markscheme

(i)     $$(1{\text{ }}3{\text{ }}6{\text{ }}5)(2{\text{ }}4)$$     A1A1

(ii)     4     A1

Note: In (b) (c) and (d) single cycles can be omitted.

[3 marks]

a.

(i)     $$(1{\text{ }}6{\text{ }}2{\text{ }}4{\text{ }}5)(3)$$     A1

(ii)     5     A1

[2 marks]

b.

$$\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 5&2&6&1&3&4 \end{array}} \right) = (1{\text{ }}5{\text{ }}3{\text{ }}6{\text{ }}4)(2)$$    (M1)A1

[2 marks]

c.

$$\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&5&2&4&6&1 \end{array}} \right) = (1{\text{ }}3{\text{ }}2{\text{ }}5{\text{ }}6)(4)$$    (M1)A1

Note: Award A2A0 for (c) and (d) combined, if answers are the wrong way round.

[2 marks]

d.

$$6! = 720$$    A2

[2 marks]

e.

any composition of the cycles (1 2), (3 4) and (5 6)     (M1)

so $${2^3} = 8$$     A1

[2 marks]

f.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

[N/A]

f.

## Question

Let $$\{ G,{\text{ }} \circ \}$$ be the group of all permutations of $$1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6$$ under the operation of composition of permutations.

Consider the following Venn diagram, where $$A = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4\} ,{\text{ }}B = \{ 3,{\text{ }}4,{\text{ }}5,{\text{ }}6\}$$.

(i)     Write the permutation $$\alpha = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&4&6&2&1&5 \end{array}} \right)$$ as a composition of disjoint cycles.

(ii)     State the order of $$\alpha$$.

[3]
a.

(i)     Write the permutation $$\beta = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 6&4&3&5&1&2 \end{array}} \right)$$ as a composition of disjoint cycles.

(ii)     State the order of $$\beta$$.

[2]
b.

Write the permutation $$\alpha \circ \beta$$ as a composition of disjoint cycles.

[2]
c.

Write the permutation $$\beta \circ \alpha$$ as a composition of disjoint cycles.

[2]
d.

State the order of $$\{ G,{\text{ }} \circ \}$$.

[2]
e.

Find the number of permutations in $$\{ G,{\text{ }} \circ \}$$ which will result in $$A$$, $$B$$ and $$A \cap B$$ remaining unchanged.

[2]
f.

## Markscheme

(i)     $$(1{\text{ }}3{\text{ }}6{\text{ }}5)(2{\text{ }}4)$$     A1A1

(ii)     4     A1

Note: In (b) (c) and (d) single cycles can be omitted.

[3 marks]

a.

(i)     $$(1{\text{ }}6{\text{ }}2{\text{ }}4{\text{ }}5)(3)$$     A1

(ii)     5     A1

[2 marks]

b.

$$\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 5&2&6&1&3&4 \end{array}} \right) = (1{\text{ }}5{\text{ }}3{\text{ }}6{\text{ }}4)(2)$$    (M1)A1

[2 marks]

c.

$$\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&5&2&4&6&1 \end{array}} \right) = (1{\text{ }}3{\text{ }}2{\text{ }}5{\text{ }}6)(4)$$    (M1)A1

Note: Award A2A0 for (c) and (d) combined, if answers are the wrong way round.

[2 marks]

d.

$$6! = 720$$    A2

[2 marks]

e.

any composition of the cycles (1 2), (3 4) and (5 6)     (M1)

so $${2^3} = 8$$     A1

[2 marks]

f.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

[N/A]

f.

## Question

Let $$\{ G,{\text{ }} \circ \}$$ be the group of all permutations of $$1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6$$ under the operation of composition of permutations.

Consider the following Venn diagram, where $$A = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4\} ,{\text{ }}B = \{ 3,{\text{ }}4,{\text{ }}5,{\text{ }}6\}$$.

(i)     Write the permutation $$\alpha = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&4&6&2&1&5 \end{array}} \right)$$ as a composition of disjoint cycles.

(ii)     State the order of $$\alpha$$.

[3]
a.

(i)     Write the permutation $$\beta = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 6&4&3&5&1&2 \end{array}} \right)$$ as a composition of disjoint cycles.

(ii)     State the order of $$\beta$$.

[2]
b.

Write the permutation $$\alpha \circ \beta$$ as a composition of disjoint cycles.

[2]
c.

Write the permutation $$\beta \circ \alpha$$ as a composition of disjoint cycles.

[2]
d.

State the order of $$\{ G,{\text{ }} \circ \}$$.

[2]
e.

Find the number of permutations in $$\{ G,{\text{ }} \circ \}$$ which will result in $$A$$, $$B$$ and $$A \cap B$$ remaining unchanged.

[2]
f.

## Markscheme

(i)     $$(1{\text{ }}3{\text{ }}6{\text{ }}5)(2{\text{ }}4)$$     A1A1

(ii)     4     A1

Note: In (b) (c) and (d) single cycles can be omitted.

[3 marks]

a.

(i)     $$(1{\text{ }}6{\text{ }}2{\text{ }}4{\text{ }}5)(3)$$     A1

(ii)     5     A1

[2 marks]

b.

$$\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 5&2&6&1&3&4 \end{array}} \right) = (1{\text{ }}5{\text{ }}3{\text{ }}6{\text{ }}4)(2)$$    (M1)A1

[2 marks]

c.

$$\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&5&2&4&6&1 \end{array}} \right) = (1{\text{ }}3{\text{ }}2{\text{ }}5{\text{ }}6)(4)$$    (M1)A1

Note: Award A2A0 for (c) and (d) combined, if answers are the wrong way round.

[2 marks]

d.

$$6! = 720$$    A2

[2 marks]

e.

any composition of the cycles (1 2), (3 4) and (5 6)     (M1)

so $${2^3} = 8$$     A1

[2 marks]

f.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

[N/A]

f.

## Question

The set of all permutations of the list of the integers 1, 2, 3  4 is a group, S4, under the operation of function composition.

In the group S4 let $${p_1} = \left( \begin{gathered} \begin{array}{*{20}{c}} 1&2&3&4 \end{array} \hfill \\ \begin{array}{*{20}{c}} 2&3&1&4 \end{array} \hfill \\ \end{gathered} \right)$$ and $${p_2} = \left( \begin{gathered} \begin{array}{*{20}{c}} 1&2&3&4 \end{array} \hfill \\ \begin{array}{*{20}{c}} 2&1&3&4 \end{array} \hfill \\ \end{gathered} \right)$$.

Determine the order of S4.

[2]
a.

Find the proper subgroup H of order 6 containing $${p_1}$$, $${p_2}$$ and their compositions. Express each element of H in cycle form.

[5]
b.

Let $$f{\text{:}}\,{S_4} \to {S_4}$$ be defined by $$f\left( p \right) = p \circ p$$ for $$p \in {S_4}$$.

Using $${p_1}$$ and $${p_2}$$, explain why $$f$$ is not a homomorphism.

[5]
c.

## Markscheme

number of possible permutations is 4 × 3 × 2 × 1       (M1)

= 24(= 4!)      A1

[2 marks]

a.

attempting to find one of $${p_1} \circ {p_1}$$, $${p_1} \circ {p_2}$$ or $${p_2} \circ {p_1}$$     M1

$${p_1} \circ {p_1} = \left( {132} \right)$$ or equivalent (eg, $${p_1}^{ – 1} = \left( {132} \right)$$)    A1

$${p_1} \circ {p_2} = \left( {13} \right)$$ or equivalent (eg, $${p_2} \circ {p_1} \circ {p_1} = \left( {13} \right)$$)    A1

$${p_2} \circ {p_1} = \left( {23} \right)$$ or equivalent (eg, $${p_1} \circ {p_1} \circ {p_2} = \left( {23} \right)$$)    A1

Note: Award A1A0A0 for one correct permutation in any form; A1A1A0 for two correct permutations in any form.

$$e = \left( 1 \right)$$, $${p_1} = \left( {123} \right)$$ and $${p_2} = \left( {12} \right)$$     A1

Note: Condone omission of identity in cycle form as long as it is clear it is considered one of the elements of H.

[5 marks]

b.

METHOD 1

if $$f$$ is a homomorphism $$f\left( {{p_1} \circ {p_2}} \right) = f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$

attempting to express one of $$f\left( {{p_1} \circ {p_2}} \right)$$ or $$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$ in terms of $${p_1}$$ and $${p_2}$$      M1

$$f\left( {{p_1} \circ {p_2}} \right) = {p_1} \circ {p_2} \circ {p_1} \circ {p_2}$$     A1

$$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right) = {p_1} \circ {p_1} \circ {p_2} \circ {p_2}$$     A1

$$\Rightarrow {p_2} \circ {p_1} = {p_1} \circ {p_2}$$     A1

but $${p_1} \circ {p_2} \ne {p_2} \circ {p_1}$$     R1

so $$f$$ is not a homomorphism     AG

Note: Award R1 only if M1 is awarded.

Note: Award marks only if $${p_1}$$ and $${p_2}$$ are used; cycle form is not required.

METHOD 2

if $$f$$ is a homomorphism $$f\left( {{p_1} \circ {p_2}} \right) = f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$

attempting to find one of $$f\left( {{p_1} \circ {p_2}} \right)$$ or $$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$      M1

$$f\left( {{p_1} \circ {p_2}} \right) = e$$     A1

$$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right) = \left( {132} \right)$$     (M1)A1

so $$f\left( {{p_1} \circ {p_2}} \right) \ne f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$     R1

so $$f$$ is not a homomorphism     AG

Note: Award R1 only if M1 is awarded.

Note: Award marks only if $${p_1}$$ and $${p_2}$$ are used; cycle form is not required.

[5 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

The group $$\{ G,{\text{ }}{ \times _7}\}$$ is defined on the set {1, 2, 3, 4, 5, 6} where $${ \times _7}$$ denotes multiplication modulo 7.

(i)     Write down the Cayley table for $$\{ G,{\text{ }}{ \times _7}\}$$ .

(ii)     Determine whether or not $$\{ G,{\text{ }}{ \times _7}\}$$ is cyclic.

(iii)     Find the subgroup of G of order 3, denoting it by H .

(iv)     Identify the element of order 2 in G and find its coset with respect to H .

[10]
a.

The group $$\{ K,{\text{ }} \circ \}$$ is defined on the six permutations of the integers 1, 2, 3 and $$\circ$$ denotes composition of permutations.

(i)     Show that $$\{ K,{\text{ }} \circ \}$$ is non-Abelian.

(ii)     Giving a reason, state whether or not $$\{ G,{\text{ }}{ \times _7}\}$$ and $$\{ K,{\text{ }} \circ \}$$ are isomorphic.

[6]
b.

## Markscheme

(i)     the Cayley table is

A3

Note: Deduct 1 mark for each error up to a maximum of 3.

(ii)     by considering powers of elements,     (M1)

it follows that 3 (or 5) is of order 6     A1

so the group is cyclic     A1

(iii)     we see that 2 and 4 are of order 3 so the subgroup of order 3 is {1, 2, 4}     M1A1

(iv)     the element of order 2 is 6     A1

the coset is {3, 5, 6}     A1

[10 marks]

a.

(i)     consider for example

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

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

Note: Award M1A1M1A0 if both compositions are done in the wrong order.

Note: Award M1A1M0A0 if the two compositions give the same result, if no further attempt is made to find two permutations which are not commutative.

these are different so the group is not Abelian     R1AG

(ii)     they are not isomorphic because $$\{ G,{\text{ }}{ \times _7}\}$$ is Abelian and $$\{ K,{\text{ }} \circ \}$$ is not     R1

[6 marks]

b.

[N/A]

a.

[N/A]

b.