| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Groups of symmetries |
| Difficulty | Challenging +1.8 This is a substantial Further Maths group theory question requiring understanding of permutation groups, geometric symmetries, Lagrange's theorem, and isomorphism. Part (a) requires interpreting cycle notation geometrically, (b) demands systematic enumeration of dihedral group D_3 elements, (c) requires proof using Lagrange's theorem and understanding of cyclic groups, and (d) requires creative construction of a shading pattern with specified symmetry properties. While the individual concepts are standard Further Maths material, the multi-part nature, need for both computational and conceptual reasoning, and the creative construction in part (d) place this well above average difficulty. |
| Spec | 8.03f Subgroups: definition and tests for proper subgroups8.03j Properties of groups: higher finite order or infinite order8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A rotation … | M1 | For identifying the permutation as representing a rotation |
| … about the centre of the shape, through an angle \(120°\) anticlockwise | A1 | Complete description including rotation, centre, angle (degrees or radians) and direction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 6 & 1 & 2 & 3 & 4 \end{pmatrix}\) | B1 | For giving the other rotation of order 3 (rotation through \(240°\) anticlockwise) |
| One of \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 6 & 5 & 4 & 3 & 2 \end{pmatrix}\), \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 6 & 5 & 4 \end{pmatrix}\) or \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 4 & 3 & 2 & 1 & 6 \end{pmatrix}\) | M1 | For one of the three reflections correctly given. NB accept cycle form for first two marks |
| Two of \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 6 & 5 & 4 & 3 & 2 \end{pmatrix}\), \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 6 & 5 & 4 \end{pmatrix}\) & \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 4 & 3 & 2 & 1 & 6 \end{pmatrix}\) | A1 | For one of the other two reflections |
| All the above and \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \end{pmatrix}\) and no extra symmetries given. Two line form required for this mark. | A1 | For all reflections and the identity, and no extra symmetries given, all in two-line notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(G\) has order 6 so can have no subgroup of order 4 by Lagrange's Theorem | B1 | Correct reason given. Stating there are no elements of order 4 is not sufficient |
| There is no element of order 6 that generates the group | B1 | Refers to there being no element of order 6 to generate the group. May reason using orders of elements or via geometric restrictions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Shape shaded to break reflection symmetries | M1 | Realises shape needs only rotations preserved; breaks reflection symmetries but preserves at least one non-trivial rotation |
| Any correctly shaded shape (e.g. examples shown) | A1 | Keeps all rotations and no reflections. Any correctly shaded shape — many variations accepted |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A rotation … | M1 | For identifying the permutation as representing a rotation |
| … about the centre of the shape, through an angle $120°$ anticlockwise | A1 | Complete description including rotation, centre, angle (degrees or radians) and direction |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 6 & 1 & 2 & 3 & 4 \end{pmatrix}$ | B1 | For giving the other rotation of order 3 (rotation through $240°$ anticlockwise) |
| One of $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 6 & 5 & 4 & 3 & 2 \end{pmatrix}$, $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 6 & 5 & 4 \end{pmatrix}$ or $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 4 & 3 & 2 & 1 & 6 \end{pmatrix}$ | M1 | For one of the three reflections correctly given. NB accept cycle form for first two marks |
| Two of $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 6 & 5 & 4 & 3 & 2 \end{pmatrix}$, $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 6 & 5 & 4 \end{pmatrix}$ & $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 4 & 3 & 2 & 1 & 6 \end{pmatrix}$ | A1 | For one of the other two reflections |
| All the above and $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \end{pmatrix}$ and no extra symmetries given. Two line form required for this mark. | A1 | For all reflections and the identity, and no extra symmetries given, all in two-line notation |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $G$ has order 6 so can have no subgroup of order 4 by Lagrange's Theorem | B1 | Correct reason given. Stating there are no elements of order 4 is **not** sufficient |
| There is no element of order 6 that generates the group | B1 | Refers to there being no element of order 6 to generate the group. May reason using orders of elements or via geometric restrictions |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Shape shaded to break reflection symmetries | M1 | Realises shape needs only rotations preserved; breaks reflection symmetries but preserves at least one non-trivial rotation |
| Any correctly shaded shape (e.g. examples shown) | A1 | Keeps all rotations and no reflections. Any correctly shaded shape — many variations accepted |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{868aedc8-6afb-4419-ae29-2ecad3461999-20_371_328_255_870}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a plane shape made up of a regular hexagon with an equilateral triangle joined to each edge and with alternate equilateral triangles shaded.
The symmetries of this shape are the rotations and reflections of the plane that preserve the shape and its shading.
The symmetries of the shape can be represented by permutations of the six vertices labelled 1 to 6 in Figure 3. The set of these permutations with the operation of composition form a group, $G$.
\begin{enumerate}[label=(\alph*)]
\item Describe geometrically the symmetry of the shape represented by the permutation
$$\left( \begin{array} { l l l l l l }
1 & 2 & 3 & 4 & 5 & 6 \\
3 & 4 & 5 & 6 & 1 & 2
\end{array} \right)$$
\item Write down, in similar two-line notation, the remaining elements of the group $G$.
\item Explain why each of the following statements is false, making your reasoning clear.
\begin{enumerate}[label=(\roman*)]
\item $G$ has a subgroup of order 4
\item $G$ is cyclic.
Diagram 1, on page 23, shows an unshaded shape with the same outline as the shape in Figure 3.
\end{enumerate}\item Shade the shape in Diagram 1 in such a way that the group of symmetries of the resulting shaded shape is isomorphic to the cyclic group of order 6
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{868aedc8-6afb-4419-ae29-2ecad3461999-23_426_378_1464_845}
\end{center}
\section*{Diagram 1}
\section*{Spare copy of Diagram 1}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{868aedc8-6afb-4419-ae29-2ecad3461999-23_424_375_2119_845}
\end{center}
Only use this diagram if you need to redraw your answer to part (d).
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2020 Q6 [10]}}