Question 5 - A-Level Maths

AQA Further Paper 3 Discrete 2021 June — Question 5 11 marks

Exam Board	AQA
Module	Further Paper 3 Discrete (Further Paper 3 Discrete)
Year	2021
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Groups
Type	Verify group axioms
Difficulty	Standard +0.3 This is a straightforward group theory question testing standard definitions and routine calculations. Parts (a) and (c) are pure recall, part (b)(i) follows immediately from commutativity of multiplication, part (b)(ii) requires computing powers of 10 mod 13 (a mechanical calculation), and part (d) involves finding subgroups using Lagrange's theorem. While this is Further Maths content, it requires no novel insight—just applying learned procedures and definitions, making it slightly easier than an average A-level question overall.
Spec	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 8.03d Latin square property: for group tables

Describe the conditions necessary for a set of elements, $S$, under a binary operation * to form a group.
5
In the multiplicative group of integers modulo 13, the group $G$ is defined as $$G = \left( \langle 10 \rangle , \times _ { 13 } \right)$$ 5 (b) (i) Explain why $G$ is an abelian group.
5 (b) (ii) Find the order of $G$.
5
State the identity element of $G$ and prove it is an identity element. Fully justify your answer.
5
Find all the proper non-trivial subgroups of $G$, giving your answers in the form $\left( \langle g \rangle , \times _ { 13 } \right)$, where $g$ is an integer less than 13

Show mark scheme Show mark scheme source

Question 5(a):

Answer	Marks	Guidance
States at least three of the four group conditions (Closure, Identity, Inverse, Associativity) correctly	B1	AO 1.1b

Closure: For all $a, b \in S$, $a*b \in S$

Identity: $\exists\, e \in S$ such that $ea = a = ae$ for all $a \in S$

Inverse: For all $a \in S$, $\exists\, b \in S$ such that $ab = e = ba$

Associativity: For all $a, b, c \in S$, $(ab)c = a(bc)$

Answer	Marks	Guidance
Describes at least two group conditions correctly	B1	AO 1.1b
Describes all four group conditions correctly	B1	AO 1.1b

Question 5(b)(i):

Answer	Marks	Guidance
$G$ is abelian as multiplication modulo $n$ is a commutative binary operation	E1	AO 2.4

Question 5(b)(ii):

Answer	Marks	Guidance
$10^1 \equiv 10 \pmod{13}$, $10^2 \equiv 9 \pmod{13}$, $10^3 \equiv 12 \pmod{13}$, $10^4 \equiv 3 \pmod{13}$, $10^5 \equiv 4 \pmod{13}$, $10^6 \equiv 1 \pmod{13}$	M1	AO 1.1a; finds and simplifies $10^2$ and $10^3$ modulo 13
Order of $G$ is 6	A1	AO 1.1b

Question 5(c):

Answer	Marks	Guidance
1 is the identity element of $G$	B1	AO 1.1b
Let $g \in G$; then $1 \times_{13} g = g$ and $g \times_{13} 1 = g$; hence both products return $g$, so 1 is the identity	M1	AO 1.1a; reasons any element multiplied by 1 remains unchanged
Completes rigorous proof using commutativity of multiplication modulo 13, abelian property, both left- and right-multiplication, and concluding statement	R1	AO 2.1

Question 5(d):

Answer	Marks	Guidance
At least one proper non-trivial subgroup of $G$: $(\langle 9 \rangle, \times_{13})$	B1	AO 1.1b; condone poor notation
Both proper non-trivial subgroups: $(\langle 9 \rangle, \times_{13})$ and $(\langle 12 \rangle, \times_{13})$ and no other subgroups	B1	AO 1.1b

## Question 5(a):

States at least three of the four group conditions (Closure, Identity, Inverse, Associativity) correctly | B1 | AO 1.1b

**Closure:** For all $a, b \in S$, $a*b \in S$
**Identity:** $\exists\, e \in S$ such that $e*a = a = a*e$ for all $a \in S$
**Inverse:** For all $a \in S$, $\exists\, b \in S$ such that $a*b = e = b*a$
**Associativity:** For all $a, b, c \in S$, $(a*b)*c = a*(b*c)$

Describes at least two group conditions correctly | B1 | AO 1.1b

Describes all four group conditions correctly | B1 | AO 1.1b

---

## Question 5(b)(i):

$G$ is abelian as multiplication modulo $n$ is a commutative binary operation | E1 | AO 2.4

---

## Question 5(b)(ii):

$10^1 \equiv 10 \pmod{13}$, $10^2 \equiv 9 \pmod{13}$, $10^3 \equiv 12 \pmod{13}$, $10^4 \equiv 3 \pmod{13}$, $10^5 \equiv 4 \pmod{13}$, $10^6 \equiv 1 \pmod{13}$ | M1 | AO 1.1a; finds and simplifies $10^2$ and $10^3$ modulo 13

Order of $G$ is 6 | A1 | AO 1.1b

---

## Question 5(c):

1 is the identity element of $G$ | B1 | AO 1.1b

Let $g \in G$; then $1 \times_{13} g = g$ and $g \times_{13} 1 = g$; hence both products return $g$, so 1 is the identity | M1 | AO 1.1a; reasons any element multiplied by 1 remains unchanged

Completes rigorous proof using commutativity of multiplication modulo 13, abelian property, both left- and right-multiplication, and concluding statement | R1 | AO 2.1

---

## Question 5(d):

At least one proper non-trivial subgroup of $G$: $(\langle 9 \rangle, \times_{13})$ | B1 | AO 1.1b; condone poor notation

Both proper non-trivial subgroups: $(\langle 9 \rangle, \times_{13})$ and $(\langle 12 \rangle, \times_{13})$ and no other subgroups | B1 | AO 1.1b

Show LaTeX source

5
\begin{enumerate}[label=(\alph*)]
\item Describe the conditions necessary for a set of elements, $S$, under a binary operation * to form a group.\\

5
\item In the multiplicative group of integers modulo 13, the group $G$ is defined as

$$G = \left( \langle 10 \rangle , \times _ { 13 } \right)$$

5 (b) (i) Explain why $G$ is an abelian group.\\

5 (b) (ii) Find the order of $G$.\\

5
\item State the identity element of $G$ and prove it is an identity element.

Fully justify your answer.\\

5
\item Find all the proper non-trivial subgroups of $G$, giving your answers in the form $\left( \langle g \rangle , \times _ { 13 } \right)$, where $g$ is an integer less than 13
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Discrete 2021 Q5 [11]}}

This paper (7 questions)

View full paper

Q1 1 Q2 1 Q3 8 Q4 7 Q5 11 Q6 8 Q7 14

AQA Further Paper 3 Discrete 2021 June — Question 5 11 marks

Question 5(a):

Closure: For all \(a, b \in S\), \(a*b \in S\)

Identity: \(\exists\, e \in S\) such that \(ea = a = ae\) for all \(a \in S\)

Inverse: For all \(a \in S\), \(\exists\, b \in S\) such that \(ab = e = ba\)

Associativity: For all \(a, b, c \in S\), \((ab)c = a(bc)\)

Question 5(b)(i):

Question 5(b)(ii):

Question 5(c):

Question 5(d):

This paper (7 questions)

Answer	Marks	Guidance
\(10^1 \equiv 10 \pmod{13}\), \(10^2 \equiv 9 \pmod{13}\), \(10^3 \equiv 12 \pmod{13}\), \(10^4 \equiv 3 \pmod{13}\), \(10^5 \equiv 4 \pmod{13}\), \(10^6 \equiv 1 \pmod{13}\)	M1	AO 1.1a; finds and simplifies \(10^2\) and \(10^3\) modulo 13
Order of \(G\) is 6	A1	AO 1.1b

Answer	Marks	Guidance
1 is the identity element of \(G\)	B1	AO 1.1b
Let \(g \in G\); then \(1 \times_{13} g = g\) and \(g \times_{13} 1 = g\); hence both products return \(g\), so 1 is the identity	M1	AO 1.1a; reasons any element multiplied by 1 remains unchanged
Completes rigorous proof using commutativity of multiplication modulo 13, abelian property, both left- and right-multiplication, and concluding statement	R1	AO 2.1

Answer	Marks	Guidance
At least one proper non-trivial subgroup of \(G\): \((\langle 9 \rangle, \times_{13})\)	B1	AO 1.1b; condone poor notation
Both proper non-trivial subgroups: \((\langle 9 \rangle, \times_{13})\) and \((\langle 12 \rangle, \times_{13})\) and no other subgroups	B1	AO 1.1b

AQA Further Paper 3 Discrete 2021 June — Question 5 11 marks

Question 5(a):

Closure: For all \(a, b \in S\), \(a*b \in S\)

Identity: \(\exists\, e \in S\) such that \(e*a = a = a*e\) for all \(a \in S\)

Inverse: For all \(a \in S\), \(\exists\, b \in S\) such that \(a*b = e = b*a\)

Associativity: For all \(a, b, c \in S\), \((a*b)*c = a*(b*c)\)

Question 5(b)(i):

Question 5(b)(ii):

Question 5(c):

Question 5(d):

This paper (7 questions)

Identity: \(\exists\, e \in S\) such that \(ea = a = ae\) for all \(a \in S\)

Inverse: For all \(a \in S\), \(\exists\, b \in S\) such that \(ab = e = ba\)

Associativity: For all \(a, b, c \in S\), \((ab)c = a(bc)\)