Question 7 - A-Level Maths

AQA Further Paper 3 Discrete 2024 June — Question 7 12 marks

Exam Board	AQA
Module	Further Paper 3 Discrete (Further Paper 3 Discrete)
Year	2024
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Groups
Type	Complete or analyse Cayley table
Difficulty	Standard +0.3 This is a standard Further Maths group theory question covering basic concepts: showing non-associativity with a counterexample, completing a Cayley table with modular arithmetic, finding inverses, determining subgroup orders using Lagrange's theorem, and identifying cyclic subgroups. All parts are routine applications of definitions and theorems with no novel insight required. While it's Further Maths content, these are foundational group theory exercises that follow textbook patterns.
Spec	8.02e Finite (modular) arithmetic: integers modulo n 8.03a Binary operations: and their properties on given sets 8.03b Cayley tables: construct for finite sets under binary operation 8.03c Group definition: recall and use, show structure is/isn't a group 8.03f Subgroups: definition and tests for proper subgroups 8.03j Properties of groups: higher finite order or infinite order

By considering associativity, show that the set of integers does not form a group under the binary operation of subtraction. Fully justify your answer. [2 marks]
The group $G$ is formed by the set $$\{1, 7, 8, 11, 12, 18\}$$ under the operation of multiplication modulo 19
1. Complete the Cayley table for $G$ [3 marks]
  $\times_{19}$ 1 7 8 11 12 18
  1 1 7 8 11 12 18
  7 7 11
  8 8 7
  11 11 7
  12 12 11
  18 18 1
2. State the inverse of 11 in $G$ [1 mark]
1. State, with a reason, the possible orders of the proper subgroups of $G$ [2 marks]
2. Find all the proper subgroups of $G$ Give your answers in the form $\langle g \rangle, \times_{19}$ where $g \in G$ [3 marks]
3. The group $H$ is such that $G \cong H$ State a possible name for $H$ [1 mark]

Show mark scheme Show mark scheme source

Question 7:

Answer	Marks
7(a)	Sets up test for associativity by

considering combinations of the

form (x – y) – z and x – (y – z),

where x, y, z , and simplifies

Answer	Marks	Guidance
one combination	1.1a	M1

3 – (2 – 1) = 2

As (3 – 2) – 1 ≠ 3 – (2 – 1), the set

of integers does not possess the

associativity property.

The associativity property is

required for all groups, therefore

the set of integers does not form a

group under subtraction.

Completes a reasoned

argument by showing a full

counterexample to associativity

(either algebraic or numerical)

and concludes that the set of

integers does not form a group

Answer	Marks	Guidance
under subtraction	2.1	R1
Subtotal	2
Q	Marking instructions	AO

Answer	Marks
7(b)(i)	Obtains at least one fully correct

row or at least one fully correct

Answer	Marks	Guidance
column of the Cayley table	1.1a	M1

Obtains at least three fully

correct rows or at least three

fully correct columns of the

Answer	Marks	Guidance
Cayley table	1.1a	M1

Completes the Cayley table

Answer	Marks	Guidance
correctly	1.1b	A1

× 1 7 8 11 12 18

1 9

1 1 7 8 11 12 18

7 7 11 18 1 8 12

8 8 18 7 12 1 11

11 11 1 12 7 18 8

12 12 8 1 18 11 7

18 18 12 11 8 7 1

Answer	Marks	Guidance
Subtotal	3
1	7	8
7	7	11

Answer	Marks	Guidance
7(b)(ii)	States 7	2.2a
Subtotal	1
Q	Marking instructions	AO

Answer	Marks
7(c)(i)	States at least two possible

correct orders for the proper

Answer	Marks	Guidance
subgroups of G	1.1a	M1

of a subgroup must divide the order

of the group.

Hence, the possible orders for the

proper subgroups of G are 1, 2 & 3

as these are divisors of 6, the order

of G

States the orders of proper

subgroups of G as 1, 2 & 3 only

(not 6) and explains that these

are divisors of 6 and names or

Answer	Marks	Guidance
explains Lagrange’s theorem	2.4	A1
Subtotal	2
Q	Marking instructions	AO

Answer	Marks
7(c)(ii)	Finds at least one correct proper

subgroup of G

Condone poor notation or

Answer	Marks	Guidance
answer given in different form	1.1a	M1

1 , ×

1 9

( )

7 , ×

1 9

( )

1 8 , ×

1 9

Finds at least two correct proper

subgroups of G

Condone poor notation or

Answer	Marks	Guidance
answer given in different form	1.1b	A1

Finds all three distinct proper

subgroups of G and no others,

giving all answers in the correct

form

( ) ( )

Note 7 , × and 1 1 , ×

1 9 1 9

Answer	Marks	Guidance
are not distinct	2.5	A1
Subtotal	3
Q	Marking instructions	AO

Answer	Marks	Guidance
7(c)(iii)	States a correct full name for H	1.2
Subtotal	1
Question total	12
Q	Marking instructions	AO

Question 7:
--- 7(a) ---
7(a) | Sets up test for associativity by
considering combinations of the
form (x – y) – z and x – (y – z),
where x, y, z , and simplifies
one combination | 1.1a | M1 | (3 – 2) – 1 = 0
3 – (2 – 1) = 2
As (3 – 2) – 1 ≠ 3 – (2 – 1), the set
of integers does not possess the
associativity property.
The associativity property is
required for all groups, therefore
the set of integers does not form a
group under subtraction.
Completes a reasoned
argument by showing a full
counterexample to associativity
(either algebraic or numerical)
and concludes that the set of
integers does not form a group
under subtraction | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b)(i) ---
7(b)(i) | Obtains at least one fully correct
row or at least one fully correct
column of the Cayley table | 1.1a | M1 | See below
Obtains at least three fully
correct rows or at least three
fully correct columns of the
Cayley table | 1.1a | M1
Completes the Cayley table
correctly | 1.1b | A1
× 1 7 8 11 12 18
1 9
1 1 7 8 11 12 18
7 7 11 18 1 8 12
8 8 18 7 12 1 11
11 11 1 12 7 18 8
12 12 8 1 18 11 7
18 18 12 11 8 7 1
Subtotal | 3
1 | 7 | 8 | 11 | 12 | 18
7 | 7 | 11 | 18 | 1 | 8 | 12
--- 7(b)(ii) ---
7(b)(ii) | States 7 | 2.2a | B1 | 7
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c)(i) ---
7(c)(i) | States at least two possible
correct orders for the proper
subgroups of G | 1.1a | M1 | By Lagrange’s theorem, the order
of a subgroup must divide the order
of the group.
Hence, the possible orders for the
proper subgroups of G are 1, 2 & 3
as these are divisors of 6, the order
of G
States the orders of proper
subgroups of G as 1, 2 & 3 only
(not 6) and explains that these
are divisors of 6 and names or
explains Lagrange’s theorem | 2.4 | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c)(ii) ---
7(c)(ii) | Finds at least one correct proper
subgroup of G
Condone poor notation or
answer given in different form | 1.1a | M1 | ( )
1 , ×
1 9
( )
7 , ×
1 9
( )
1 8 , ×
1 9
Finds at least two correct proper
subgroups of G
Condone poor notation or
answer given in different form | 1.1b | A1
Finds all three distinct proper
subgroups of G and no others,
giving all answers in the correct
form
( ) ( )
Note 7 , × and 1 1 , ×
1 9 1 9
are not distinct | 2.5 | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c)(iii) ---
7(c)(iii) | States a correct full name for H | 1.2 | B1 | Cyclic group of order 6
Subtotal | 1
Question total | 12
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

\begin{enumerate}[label=(\alph*)]
\item By considering associativity, show that the set of integers does not form a group under the binary operation of subtraction.

Fully justify your answer.
[2 marks]

\item The group $G$ is formed by the set
$$\{1, 7, 8, 11, 12, 18\}$$
under the operation of multiplication modulo 19

\begin{enumerate}[label=(\roman*)]
\item Complete the Cayley table for $G$
[3 marks]

\begin{tabular}{c|c|c|c|c|c|c}
$\times_{19}$ & 1 & 7 & 8 & 11 & 12 & 18 \\
\hline
1 & 1 & 7 & 8 & 11 & 12 & 18 \\
7 & 7 & 11 & & & & \\
8 & 8 & & 7 & & & \\
11 & 11 & & & 7 & & \\
12 & 12 & & & & 11 & \\
18 & 18 & & & & & 1 \\
\end{tabular}

\item State the inverse of 11 in $G$
[1 mark]
\end{enumerate}

\item \begin{enumerate}[label=(\roman*)]
\item State, with a reason, the possible orders of the proper subgroups of $G$
[2 marks]

\item Find all the proper subgroups of $G$

Give your answers in the form $\langle g \rangle, \times_{19}$ where $g \in G$
[3 marks]

\item The group $H$ is such that $G \cong H$

State a possible name for $H$
[1 mark]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Discrete 2024 Q7 [12]}}

This paper (10 questions)

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Q1 1 Q2 1 Q3 1 Q4 4 Q5 4 Q6 6 Q7 12 Q8 8 Q9 6 Q10 7

\(\times_{19}\)	1	7	8	11	12	18
1	1	7	8	11	12	18
7	7	11
8	8		7
11	11			7
12	12				11
18	18					1