| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Matrix groups |
| Difficulty | Challenging +1.2 This is a structured group theory question with clear steps: computing matrix products mod 2 (routine calculation), finding element orders (systematic but straightforward), identifying non-isomorphism via order structure (standard reasoning), and finding an explicit isomorphism between two small groups (requires matching elements but both groups are given completely). While it covers multiple group theory concepts, each part follows standard techniques without requiring novel insight. It's moderately above average due to the Further Maths content and multi-part nature, but remains accessible to well-prepared students. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{B}^2 = \begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}1&1\\1&0\end{pmatrix} = \begin{pmatrix}0&1\\1&1\end{pmatrix}\{=\mathbf{E}\}\) or \(\mathbf{B}^3 = \begin{pmatrix}3&2\\2&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}\) | M1 | 2.5 |
| \(\mathbf{B}^3 = \begin{pmatrix}0&1\\1&1\end{pmatrix}\begin{pmatrix}1&1\\1&0\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}\) and B is not the identity, hence B has order 3 or \( | \mathbf{B} | =3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| I has order 1, and E (which is \(\mathbf{B}^{-1}\)) has order 3 | B1 | 1.1b |
| Finds order of at least one remaining element, e.g. \(\mathbf{A}^2 = \begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}\) or \(\mathbf{C}^2 = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}\) or \(\mathbf{D}^2 = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}\) | M1 | 1.1b |
| So I has order 1, B and E order 3, and A, C and D each have order 2 | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) There is no element of order 6 in \(G\), so cannot be isomorphic to \(C_6\) | B1 | 2.4 |
| (ii) Group of symmetries of a regular hexagon has more than 6 elements / has 12 symmetries / has 2 rotations of order 6 / has 2 rotations of order 3 / has 1 rotation of order 2 / has 6 reflections of order 2 / order greater than 6 | B1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Any two correct matchings | M1 | 2.2a |
| Completes matching with at least 4 correct | M1 | 3.1a |
| Fully specified correct isomorphism, e.g. \(\mathbf{I}\to e,\ \mathbf{A}\to c,\ \mathbf{B}\to a,\ \mathbf{C}\to d,\ \mathbf{D}\to f,\ \mathbf{E}\to b\) | A1 | 2.1 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{B}^2 = \begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}1&1\\1&0\end{pmatrix} = \begin{pmatrix}0&1\\1&1\end{pmatrix}\{=\mathbf{E}\}$ or $\mathbf{B}^3 = \begin{pmatrix}3&2\\2&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}$ | **M1** | 2.5 |
| $\mathbf{B}^3 = \begin{pmatrix}0&1\\1&1\end{pmatrix}\begin{pmatrix}1&1\\1&0\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}$ and **B** is not the identity, hence **B** has order 3 or $|\mathbf{B}|=3$ | **A1** | 2.1 |
**Notes (a):**
- **M1:** Squares **B** to begin finding the order, or uses calculator to find $\mathbf{B}^3$
- **A1:** Shows $\mathbf{B}^3 = \mathbf{I}$ and concludes order must be 3
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **I** has order 1, and **E** (which is $\mathbf{B}^{-1}$) has order 3 | **B1** | 1.1b |
| Finds order of at least one remaining element, e.g. $\mathbf{A}^2 = \begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}$ or $\mathbf{C}^2 = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}$ or $\mathbf{D}^2 = \begin{pmatrix}1&0\\0&1\end{pmatrix}\{=\mathbf{I}\}$ | **M1** | 1.1b |
| So **I** has order 1, **B** and **E** order 3, and **A**, **C** and **D** each have order 2 | **A1** | 1.1b |
**Notes (b):**
- **B1:** States order for **I** and gives order for **E** (inverse of **B**)
- **M1:** Finds order of at least one other element to get element of order 2
- **A1:** Correct orders for all three elements of order 2
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) There is no element of **order 6** in $G$, so cannot be isomorphic to $C_6$ | **B1** | 2.4 |
| (ii) Group of symmetries of a regular hexagon has more than 6 elements / has 12 symmetries / has 2 rotations of order 6 / has 2 rotations of order 3 / has 1 rotation of order 2 / has 6 reflections of order 2 / order greater than 6 | **B1** | 2.4 |
**Notes (c):**
- **B1:** Correct reason — identifies no element of **order 6**. Not sufficient to say it has no generator
- **B1:** Any correct reason given. B0 for: group of symmetries also cyclic; groups not same order; incorrect orders stated; comments only on order of whole group; states symmetry group has order 6
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Any two correct matchings | **M1** | 2.2a |
| Completes matching with at least 4 correct | **M1** | 3.1a |
| Fully specified correct isomorphism, e.g. $\mathbf{I}\to e,\ \mathbf{A}\to c,\ \mathbf{B}\to a,\ \mathbf{C}\to d,\ \mathbf{D}\to f,\ \mathbf{E}\to b$ | **A1** | 2.1 |
**Notes (d):**
- **M1:** Any two correct matchings
- **M1:** Completes matching with at least 4 correct
- **A1:** A fully specified correct isomorphism
---\begin{enumerate}
\item The set of matrices $G = \{ \mathbf { I } , \mathbf { A } , \mathbf { B } , \mathbf { C } , \mathbf { D } , \mathbf { E } \}$ where
\end{enumerate}
$$\mathbf { I } = \left( \begin{array} { l l }
1 & 0 \\
0 & 1
\end{array} \right) \quad \mathbf { A } = \left( \begin{array} { l l }
0 & 1 \\
1 & 0
\end{array} \right) \quad \mathbf { B } = \left( \begin{array} { l l }
1 & 1 \\
1 & 0
\end{array} \right) \quad \mathbf { C } = \left( \begin{array} { l l }
1 & 1 \\
0 & 1
\end{array} \right) \quad \mathbf { D } = \left( \begin{array} { l l }
1 & 0 \\
1 & 1
\end{array} \right) \quad \mathbf { E } = \left( \begin{array} { l l }
0 & 1 \\
1 & 1
\end{array} \right)$$
with the operation $\otimes _ { 2 }$ of matrix multiplication with entries evaluated modulo 2 , forms a group.\\
(a) Show that $\mathbf { B }$ is an element of order 3 in $G$.\\
(b) Determine the orders of the other elements of $G$.\\
(c) Give a reason why $G$ is not isomorphic to\\
(i) a cyclic group of order 6\\
(ii) the group of symmetries of a regular hexagon.
The group $H$ of permutations of the numbers 1, 2 and 3 contains the following elements, denoted in two-line notation,
$$\begin{array} { l l l }
e = \left( \begin{array} { l l l }
1 & 2 & 3 \\
1 & 2 & 3
\end{array} \right) & a = \left( \begin{array} { l l l }
1 & 2 & 3 \\
2 & 3 & 1
\end{array} \right) & b = \left( \begin{array} { l l l }
1 & 2 & 3 \\
3 & 1 & 2
\end{array} \right) \\
c = \left( \begin{array} { l l }
1 & 2 \\
1 & 3 \\
2
\end{array} \right) & d = \left( \begin{array} { l l l }
1 & 2 & 3 \\
2 & 1 & 3
\end{array} \right) & f = \left( \begin{array} { l l }
1 & 2 \\
3 & 2
\end{array} \right)
\end{array}$$
(d) Determine an isomorphism between the groups $G$ and $H$.
\hfill \mbox{\textit{Edexcel FP2 2024 Q7 [10]}}