| Exam Board | OCR MEI |
|---|---|
| Module | Further Extra Pure (Further Extra Pure) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Complete or analyse Cayley table |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring abstract reasoning about group axioms rather than computation. Students must deduce properties from minimal information (a∘a=2 for all a), apply closure/identity/inverse axioms systematically, and construct the complete group table through logical deduction. While methodical, it demands sophisticated mathematical maturity beyond standard A-level and involves non-trivial problem-solving rather than routine application. |
| Spec | 8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | 2·2=2 |
| so 2 must be the identity | B1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 2.2a | Allow other fully |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (b) | 1·3=1 would mean that 3 is the identity, and |
| Answer | Marks |
|---|---|
| hence 1·3=4 | B1 |
| Answer | Marks |
|---|---|
| B1 | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | oe, via 1·3=3·3Þ1=3 |
| Answer | Marks |
|---|---|
| [3] | B1 for 4 only |
| Answer | Marks |
|---|---|
| Explicitly stated | Min requirements: |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (c) | (i) |
| Answer | Marks |
|---|---|
| 4 3 4 1 2 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | (Values in shaded cells given in |
| Answer | Marks |
|---|---|
| All correct | Diagonal wrong M0 |
| Answer | Marks | Guidance |
|---|---|---|
| | 1 | 2 |
| 4 | 3 | 4 |
| 4 | (c) | (ii) |
| leading diagonal (oe) | B1ft | 1.2 |
| justification | Must follow from their |
| Answer | Marks |
|---|---|
| Both groups of order 4 are abelian | 1.2 |
Question 4:
4 | (a) | 2·2=2
so 2 must be the identity | B1
B1
[2] | 2.2a
2.2a | Allow other fully
correct explanations
4 | (b) | 1·3=1 would mean that 3 is the identity, and
1·3=3 would mean that 1 is the identity
1·3=2 implies 1·3=1·1Þ3=1, so
contradiction
hence 1·3=4 | B1
B1
B1 | 3.1a
2.4
2.2a | oe, via 1·3=3·3Þ1=3
B1 for 4 only
B1
B1
B1
[3] | B1 for 4 only
No repetition in row/column
Explicitly stated | Min requirements:
1.1=2 or 3.3=2
1.2=1 and 2.3=3
Alternative Solution:
From the (partially) constructed table
Leading diagonal
Identity row and column
1∙3 = 4
4 | (c) | (i) | 1 2 3 4
1 2 1 4 3
2 1 2 3 4
3 4 3 2 1
4 3 4 1 2 | M1
A1
[2] | 1.1
1.1 | (Values in shaded cells given in
part a)
M1 for any 7 new entries that are
not repeated in the row and
column.
All correct | Diagonal wrong M0
B1
B1
B1
[3]
B1 for 4 only
No repetition in row/column
Explicitly stated
Min requirements:
1.1=2 or 3.3=2
1.2=1 and 2.3=3
| 1 | 2 | 3 | 4
4 | 3 | 4 | 1 | 2
4 | (c) | (ii) | Group is abelian as table is symmetrical about the
leading diagonal (oe) | B1ft | 1.2 | Correct conclusion with
justification | Must follow from their
complete table
Alternative Solution:
Both groups of order 4 are abelian | 1.2
B1
[1]
Alternative Solution:
Both groups of order 4 are abelian
1.2$T$ is the set $\{1, 2, 3, 4\}$. A binary operation $\bullet$ is defined on $T$ such that $a \bullet a = 2$ for all $a \in T$. It is given that $(T, \bullet)$ is a group.
\begin{enumerate}[label=(\alph*)]
\item Deduce the identity element in $T$, giving a reason for your answer. [2]
\item Find the value of $1 \bullet 3$, showing how the result is obtained. [3]
\item \begin{enumerate}[label=(\roman*)]
\item Complete a group table for $(T, \bullet)$. [2]
\item State with a reason whether the group is abelian. [1]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Extra Pure 2019 Q4 [8]}}