| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Complete or analyse Cayley table |
| Difficulty | Challenging +1.2 This is a Further Maths group theory question requiring systematic matrix multiplication to complete a Cayley table, then standard group-theoretic analysis. While the topic is advanced, the execution is methodical: multiply given 2×2 matrices, identify patterns, find inverses/orders from the table, and apply Lagrange's theorem. The quaternion group structure and non-isomorphism argument require some insight but follow standard FP3 techniques. |
| Spec | 8.03b Cayley tables: construct for finite sets under binary operation8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups |
| I | J | K | L | -I | -J | -K | \(- \mathbf { L }\) | |
| I | I | J | K | L | -I | -J | -K | -L |
| J | J | -I | L | -K | -J | I | -L | K |
| K | K | -L | -I | |||||
| L | L | K | ||||||
| -I | -I | -J | ||||||
| -J | -J | I | ||||||
| -K | -K | L | ||||||
| -L | -L | -K |
| Answer | Marks | Guidance |
|---|---|---|
| Complete Cayley table using \(\mathbf{JK=L}\), \(\mathbf{KJ=-L}\) and matrix multiplication rules | M1 M1 M1 M1 M1 A1 | Method marks for systematic use of relations; A1 for fully correct table |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{I}^{-1}=\mathbf{I}\); \((-\mathbf{I})^{-1}=-\mathbf{I}\) | B1 | |
| \(\mathbf{J}^{-1}=-\mathbf{J}\); \((-\mathbf{J})^{-1}=\mathbf{J}\) | B1 | |
| \(\mathbf{K}^{-1}=-\mathbf{K}\); \(\mathbf{L}^{-1}=-\mathbf{L}\) and negatives similarly | B1 | All correct |
| Answer | Marks |
|---|---|
| Order of \(\mathbf{I}\): 1 | B1 |
| Order of \(-\mathbf{I}\): 2 (since \((-\mathbf{I})^2=\mathbf{I}\)) | B1 |
| Orders of \(\mathbf{J},-\mathbf{J},\mathbf{K},-\mathbf{K},\mathbf{L},-\mathbf{L}\): all 4 | B1 |
| Answer | Marks |
|---|---|
| By Lagrange's theorem, order of subgroup divides order of group (8) | B1 |
| A subgroup of order 4 has index 2, hence is normal | B1 |
| Every group of order 4 is either cyclic (\(\mathbb{Z}_4\)) or Klein 4-group; since all non-identity elements except \(-\mathbf{I}\) have order 4, any order-4 subgroup contains an element of order 4, hence must be cyclic | B1 B1 |
| Answer | Marks |
|---|---|
| Proper subgroups of order 2: \(\{\mathbf{I},-\mathbf{I}\}\) | B1 |
| Proper subgroups of order 4: \(\{\mathbf{I},-\mathbf{I},\mathbf{J},-\mathbf{J}\}\), \(\{\mathbf{I},-\mathbf{I},\mathbf{K},-\mathbf{K}\}\), \(\{\mathbf{I},-\mathbf{I},\mathbf{L},-\mathbf{L}\}\) | B1 B1 B1 |
| Trivial subgroup \(\{\mathbf{I}\}\) | B1 |
| Answer | Marks |
|---|---|
| Group of symmetries of square has an element of order 2 (e.g. reflection) that is not the unique element of order 2 | M1 |
| In \(G\), there is only one element of order 2, namely \(-\mathbf{I}\) | A1 |
| In symmetry group of square, there are multiple elements of order 2; hence not isomorphic | A1 |
# Question 4: (Option 4: Groups)
## Part (i)
| Complete Cayley table using $\mathbf{JK=L}$, $\mathbf{KJ=-L}$ and matrix multiplication rules | M1 M1 M1 M1 M1 A1 | Method marks for systematic use of relations; A1 for fully correct table |
Key entries include:
- $\mathbf{KK=-I}$, $\mathbf{KL=J}$ (row K)
- $\mathbf{LK=-J}$, $\mathbf{LL=-I}$ (row L)
- Rows $-\mathbf{I},-\mathbf{J},-\mathbf{K},-\mathbf{L}$ are negatives of $\mathbf{I,J,K,L}$ rows
## Part (ii)
| $\mathbf{I}^{-1}=\mathbf{I}$; $(-\mathbf{I})^{-1}=-\mathbf{I}$ | B1 | |
|---|---|---|
| $\mathbf{J}^{-1}=-\mathbf{J}$; $(-\mathbf{J})^{-1}=\mathbf{J}$ | B1 | |
| $\mathbf{K}^{-1}=-\mathbf{K}$; $\mathbf{L}^{-1}=-\mathbf{L}$ and negatives similarly | B1 | All correct |
## Part (iii)
| Order of $\mathbf{I}$: 1 | B1 | |
|---|---|---|
| Order of $-\mathbf{I}$: 2 (since $(-\mathbf{I})^2=\mathbf{I}$) | B1 | |
| Orders of $\mathbf{J},-\mathbf{J},\mathbf{K},-\mathbf{K},\mathbf{L},-\mathbf{L}$: all 4 | B1 | |
## Part (iv)
| By Lagrange's theorem, order of subgroup divides order of group (8) | B1 | |
|---|---|---|
| A subgroup of order 4 has index 2, hence is normal | B1 | |
| Every group of order 4 is either cyclic ($\mathbb{Z}_4$) or Klein 4-group; since all non-identity elements except $-\mathbf{I}$ have order 4, any order-4 subgroup contains an element of order 4, hence must be cyclic | B1 B1 | |
## Part (v)
| Proper subgroups of order 2: $\{\mathbf{I},-\mathbf{I}\}$ | B1 | |
|---|---|---|
| Proper subgroups of order 4: $\{\mathbf{I},-\mathbf{I},\mathbf{J},-\mathbf{J}\}$, $\{\mathbf{I},-\mathbf{I},\mathbf{K},-\mathbf{K}\}$, $\{\mathbf{I},-\mathbf{I},\mathbf{L},-\mathbf{L}\}$ | B1 B1 B1 | |
| Trivial subgroup $\{\mathbf{I}\}$ | B1 | |
## Part (vi)
| Group of symmetries of square has an element of order 2 (e.g. reflection) that is not the unique element of order 2 | M1 | |
|---|---|---|
| In $G$, there is only one element of order 2, namely $-\mathbf{I}$ | A1 | |
| In symmetry group of square, there are multiple elements of order 2; hence not isomorphic | A1 | |
---$\mathbf { 4 }$ The group $G$ consists of the 8 complex matrices $\{ \mathbf { I } , \mathbf { J } , \mathbf { K } , \mathbf { L } , - \mathbf { I } , - \mathbf { J } , - \mathbf { K } , - \mathbf { L } \}$ under matrix multiplication, where
$$\mathbf { I } = \left( \begin{array} { l l }
1 & 0 \\
0 & 1
\end{array} \right) , \quad \mathbf { J } = \left( \begin{array} { r r }
\mathrm { j } & 0 \\
0 & - \mathrm { j }
\end{array} \right) , \quad \mathbf { K } = \left( \begin{array} { r r }
0 & 1 \\
- 1 & 0
\end{array} \right) , \quad \mathbf { L } = \left( \begin{array} { c c }
0 & \mathrm { j } \\
\mathrm { j } & 0
\end{array} \right)$$
(i) Copy and complete the following composition table for $G$.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
& I & J & K & L & -I & -J & -K & $- \mathbf { L }$ \\
\hline
I & I & J & K & L & -I & -J & -K & -L \\
\hline
J & J & -I & L & -K & -J & I & -L & K \\
\hline
K & K & -L & -I & & & & & \\
\hline
L & L & K & & & & & & \\
\hline
-I & -I & -J & & & & & & \\
\hline
-J & -J & I & & & & & & \\
\hline
-K & -K & L & & & & & & \\
\hline
-L & -L & -K & & & & & & \\
\hline
\end{tabular}
\end{center}
(Note that $\mathbf { J K } = \mathbf { L }$ and $\mathbf { K J } = - \mathbf { L }$.)\\
(ii) State the inverse of each element of $G$.\\
(iii) Find the order of each element of $G$.\\
(iv) Explain why, if $G$ has a subgroup of order 4, that subgroup must be cyclic.\\
(v) Find all the proper subgroups of $G$.\\
(vi) Show that $G$ is not isomorphic to the group of symmetries of a square.
\hfill \mbox{\textit{OCR MEI FP3 2006 Q4 [24]}}