| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Matrix groups |
| Difficulty | Challenging +1.2 This is a Further Maths group theory question with multiple parts, but most are straightforward applications of definitions. Part (a) is trivial recall, (b) requires recognizing a rotation matrix and computing powers until returning to identity, (c) involves systematic enumeration of a small group, and (d) tests understanding of basic group properties. While the topic is advanced, the actual problem-solving required is mechanical rather than requiring deep insight. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups8.03g Cyclic groups: meaning of the term8.03h Generators: of cyclic and non-cyclic groups |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 7 1 1 9 = 9 + (1 23) + (1 232) + (7 233) |
| Answer | Marks |
|---|---|
| 10 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 1.1 | Clear use of base-23 column-values | |
| 5 | (b) | 7n + 11 9 (mod 23) 7n + 11 32 (mod 23) |
| Answer | Marks |
|---|---|
| hcf(7, 23) = 1 for division to be valid | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | Use of modular arithmetic to gain a multiple of 7 on |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | i |
| Answer | Marks |
|---|---|
| hcf(7, 23) = 1 and hcf(3, 23) = 1 | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | Must be written explicitly as a multiple of 23, here or |
Question 5:
5 | (a) | 7 1 1 9 = 9 + (1 23) + (1 232) + (7 233)
2 3
= 85 730
10 | M1
A1
[2] | 1.1
1.1 | Clear use of base-23 column-values
5 | (b) | 7n + 11 9 (mod 23) 7n + 11 32 (mod 23)
OR 7n −2 (mod 23)
7n 21 (mod 23)
n 3 (mod 23) or n = 23k + 3 (k ℤ)
hcf(7, 23) = 1 for division to be valid | M1
A1
B1
[3] | 1.1
1.1
2.4 | Use of modular arithmetic to gain a multiple of 7 on
the RHS
Any correct, complete statement
Accept n −20 (mod 23)
Statement that hcf(7, 23) = 1 for division to be valid
or equivalent
5 | (c) | i | 3N – 7M = 30a + 3b – 7a – 49b = 23(a – 2b)
If N = 23k, then 7M = 23(3k – a + 2b)
If M = 23k, then 3N = 23(7k + a – 2b)
Both correctly shown multiples of 23 with explanation that
hcf(7, 23) = 1 and hcf(3, 23) = 1 | B1
M1
M1
A1
[4] | 1.1
2.1
2.1
2.4 | Must be written explicitly as a multiple of 23, here or
later on
Proof attempted in one direction
(“proof” includes attempt to obtain a multiple of 23)
Proof attempted in other direction
(“proof” includes attempt to obtain a multiple of 23)
At least one justification must be noted
[2]
5 The set $S$ consists of all $2 \times 2$ matrices having determinant 1 or - 1 . For instance, the matrices $\mathbf { P } = \left( \begin{array} { c c } \frac { 1 } { 2 } & \frac { \sqrt { 3 } } { 2 } \\ - \frac { \sqrt { 3 } } { 2 } & \frac { 1 } { 2 } \end{array} \right) , \mathbf { Q } = - \left( \begin{array} { c c } \frac { 1 } { 2 } & \frac { \sqrt { 3 } } { 2 } \\ - \frac { \sqrt { 3 } } { 2 } & \frac { 1 } { 2 } \end{array} \right)$ and $\mathbf { R } = \left( \begin{array} { r r } 1 & 0 \\ 0 & - 1 \end{array} \right)$ are elements of $S$. It is given that $\times _ { \mathbf { M } }$ is the operation of matrix multiplication.
\begin{enumerate}[label=(\alph*)]
\item State the identity element of $S$ under $\times _ { \mathbf { M } }$.
The group $G$ is generated by $\mathbf { P }$, under $\times _ { \mathbf { M } }$.
\item Determine the order of $G$.
The group $H$ is generated by $\mathbf { Q }$ and $\mathbf { R }$, also under $\times _ { \mathbf { M } }$.
\item \begin{enumerate}[label=(\roman*)]
\item By finding each element of $H$, determine the order of $H$.
\item List all the proper subgroups of $H$.
\end{enumerate}\item State whether each of the following statements is true or false. Give a reason for each of your answers.
\begin{itemize}
\item $G$ is abelian
\item $G$ is cyclic
\item $H$ is abelian
\item $H$ is cyclic
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2024 Q5 [14]}}