| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Discrete (Further Paper 3 Discrete) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Isomorphism between groups |
| Difficulty | Challenging +1.2 This is a Further Maths group theory question requiring students to verify a generator, prove isomorphism between cyclic groups of order 4, and find subgroups. While it involves abstract algebra concepts beyond standard A-level, the actual tasks are fairly routine: computing matrix powers to show B generates G, defining an explicit isomorphism Ļ:GāH, applying Lagrange's theorem, and listing subgroups. The question is structured with clear guidance and requires standard techniques rather than novel insight, making it moderately above average difficulty but accessible to well-prepared Further Maths students. |
| Spec | 8.03f Subgroups: definition and tests for proper subgroups8.03g Cyclic groups: meaning of the term8.03h Generators: of cyclic and non-cyclic groups8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}^2 = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix} = C\) | M1 | Finds correctly for \(n = 2\), 3 or 4 |
| \(\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}^3 = \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix} = D\); \(\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}^4 = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} = A\) | A1 | Finds correctly for \(n = 2\), 3 and 4 |
| As each of \(B, B^2, B^3, B^4\) is a different element of \(G\), then \(B = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}\) is a generator of \(G\) | R1 | Concludes repeated multiplication produces every element of \(G\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(B = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix} \mapsto \mathrm{i}\) (at least one correct mapping stated, or states \(G\) is cyclic of order 4) | B1 | Condone poor notation |
| \(B^2 = C = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix} \mapsto -1 = \mathrm{i}^2\); \(B^3 = D = \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix} \mapsto -\mathrm{i} = \mathrm{i}^3\); \(B^4 = A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \mapsto 1 = \mathrm{i}^4\) | B1 | All correct mappings with correct notation; or identifies \(\mathrm{i}\) as generator of \(H\) |
| Both groups have the same structure preserved by the one-to-one mapping; \(G\) is isomorphic to \(H\) | E1 | Deduces one-to-one mapping between each element of \(G\) and \(H\); or deduces \(H\) is cyclic of order 4 |
| Completes rigorous argument to conclude \(G\) is isomorphic to \(H\) | R1 |
| Answer | Marks |
|---|---|
| The order of \(H\) is 4 | B1 |
| 4 is not divisible by 3, so \(H\) cannot have a subgroup of order 3 by Lagrange's theorem | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{1,\ -1\}\) | M1 | One subgroup found; condone poor notation |
| \(\{1\}\) | A1 | Second subgroup found |
| \(H = \{1,\ -1,\ \mathrm{i},\ -\mathrm{i}\}\) ā all three subgroups and no others, including \(H\) itself and the trivial subgroup | A1 | All three correct with no extras, correct notation |
## Question 5(a)(i):
| $\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}^2 = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix} = C$ | M1 | Finds correctly for $n = 2$, 3 or 4 |
| $\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}^3 = \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix} = D$; $\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}^4 = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} = A$ | A1 | Finds correctly for $n = 2$, 3 and 4 |
| As each of $B, B^2, B^3, B^4$ is a different element of $G$, then $B = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$ is a generator of $G$ | R1 | Concludes repeated multiplication produces every element of $G$ |
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## Question 5(a)(ii):
| $B = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix} \mapsto \mathrm{i}$ (at least one correct mapping stated, or states $G$ is cyclic of order 4) | B1 | Condone poor notation |
| $B^2 = C = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix} \mapsto -1 = \mathrm{i}^2$; $B^3 = D = \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix} \mapsto -\mathrm{i} = \mathrm{i}^3$; $B^4 = A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \mapsto 1 = \mathrm{i}^4$ | B1 | All correct mappings with correct notation; or identifies $\mathrm{i}$ as generator of $H$ |
| Both groups have the same structure preserved by the one-to-one mapping; $G$ is isomorphic to $H$ | E1 | Deduces one-to-one mapping between each element of $G$ and $H$; or deduces $H$ is cyclic of order 4 |
| Completes rigorous argument to conclude $G$ is isomorphic to $H$ | R1 | |
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## Question 5(b)(i):
| The order of $H$ is 4 | B1 | |
| 4 is not divisible by 3, so $H$ cannot have a subgroup of order 3 by Lagrange's theorem | E1 | |
---
## Question 5(b)(ii):
| $\{1,\ -1\}$ | M1 | One subgroup found; condone poor notation |
| $\{1\}$ | A1 | Second subgroup found |
| $H = \{1,\ -1,\ \mathrm{i},\ -\mathrm{i}\}$ ā all three subgroups and no others, including $H$ itself and the trivial subgroup | A1 | All three correct with no extras, correct notation |
---5 The set $S$ is defined as
$$S = \{ A , B , C , D \}$$
where\\
$A = \left[ \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right] \quad B = \left[ \begin{array} { c c } 0 & - 1 \\ 1 & 0 \end{array} \right] \quad C = \left[ \begin{array} { c c } - 1 & 0 \\ 0 & - 1 \end{array} \right] \quad D = \left[ \begin{array} { c c } 0 & 1 \\ - 1 & 0 \end{array} \right]$
The group $G$ is formed by $S$ under matrix multiplication.\\
The group $H$ is defined as $H = ( \langle \mathrm { i } \rangle , \times )$, where $\mathrm { i } ^ { 2 } = - 1$\\
5
\begin{enumerate}[label=(\alph*)]
\item (i) Prove that $B$ is a generator of $G$.\\
Fully justify your answer.\\
5 (a) (ii) Show that $G \cong H$.\\
Fully justify your answer.\\
5
\item (i) Explain why $H$ has no subgroups of order 3\\
Fully justify your answer.\\
5 (b) (ii) Find all of the subgroups of $H$.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Discrete 2019 Q5 [12]}}