| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Groups with generators and relations |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring multiple proofs about group structure, element orders, and subgroup verification. While it involves abstract algebra concepts beyond standard A-level, the question provides significant scaffolding through the given properties, and the proofs follow relatively standard manipulations once students recognize how to use a² = p² = (ap)². The multi-part structure and need for formal proof reasoning place it well above average difficulty, but it's more systematic than genuinely novel. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03d Latin square property: for group tables8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups8.03k Lagrange's theorem: order of subgroup divides order of group |
| Answer | Marks | Guidance |
|---|---|---|
| \(a^2 = (ap)^2 = apap \Rightarrow a = pap\) | B1 | For use of given properties to obtain AG |
| \(p^2 = (ap)^2 = apap \Rightarrow p = apa\) | B1, 2 | For use of given properties to obtain AG. SR allow working from AG to obtain relevant properties |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(p^2\right)^2 = p^4 = e \Rightarrow\) order \(p^2 = 2\) | B1 | For correct order with no incorrect working seen |
| \(\left(a^2\right)^2 = \left(p^2\right)^2 = e \Rightarrow\) order \(a = 4\) | B1 | For correct order with no incorrect working seen |
| \((ap)^4 = a^4 = e \Rightarrow\) order \(ap = 4\) | B1 | For correct order with no incorrect working seen |
| \(\left(ap^2\right)^2 = ap^2ap^2 = ap \cdot a \cdot p = a^2\) | M1 | For relevant use of (i) or given properties |
| OR \(ap^2 = a \cdot a^2 = a^3 \Rightarrow\) | A1, 5 | For correct order with no incorrect working seen |
| \(\left(ap^2\right)^2 = a^6 = a^2\) | ||
| \(\Rightarrow\) order \(ap^2 = 4\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p^2 = a^2, ap^2 = a^3\) | M2 | For use of the given properties to simplify \(p^2\) and \(ap^2\) |
| \(\Rightarrow [e, a, p^2, ap^2] = [e, a, a^2, a^3]\) | A1 | For obtaining \(a^2\) and \(a^3\) |
| which is a cyclic group | A1, 4 | For justifying that the set is a group |
| Answer | Marks | Guidance |
|---|---|---|
| \(e\) | \(a\) | \(p^2\) |
| \(e\) | \(e\) | \(a\) |
| \(a\) | \(a\) | \(p^2\) |
| \(p^2\) | \(p^2\) | \(ap^2\) |
| \(ap^2\) | \(ap^2\) | \(e\) |
| M1, A1 | For attempting closure with all 9 non-trivial products seen. For all 16 products correct | |
| Completed table is a cyclic group | B2 | For justifying that the set is a group |
| Answer | Marks | Guidance |
|---|---|---|
| \(e\) | \(a\) | \(p^2\) |
| \(e\) | \(e\) | \(a\) |
| \(a\) | \(a\) | \(p^2\) |
| \(p^2\) | \(p^2\) | \(ap^2\) |
| \(ap^2\) | \(ap^2\) | \(e\) |
| M1, A1 | For attempting closure with all 9 non-trivial products seen. For all 16 products correct | |
| Identity = \(e\) | B1 | For stating identity |
| Inverses exist since EITHER: \(e\) is in each row/column OR: \(p^2\) is self-inverse; \(a, ap^2\) form an inverse pair | B1 | For justifying inverses (\(e^{-1} = e\) may be assumed) |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. \(a \cdot ap = a^2p = p^3\) } \(\Rightarrow\) not | M1 | For attempting to find a non-commutative pair of elements, at least one involving \(a\) (may be embedded in a full or partial table) |
| \(ap \cdot a = p\) | ||
| M1, B1, A1, 4 | For simplifying elements both ways round. For a correct pair of non-commutative elements. For stating \(Q\) non-commutative, with a clear argument |
| Answer | Marks | Guidance |
|---|---|---|
| Assume commutativity, so (eg) \(ap = pa\) | M1 | For setting up proof by contradiction |
| (i) \(\Rightarrow p = ap.a \Rightarrow pa.a = pa^2 = pp^2 = p^3\) | M1 | For using (i) and/or given properties |
| But \(p\) and \(p^3\) are distinct | B1 | For obtaining and stating a contradiction |
| \(\Rightarrow Q\) is non-commutative | A1 | For stating \(Q\) non-commutative, with a clear argument |
**Part (i)**
| $a^2 = (ap)^2 = apap \Rightarrow a = pap$ | B1 | For use of given properties to obtain AG |
|---|---|---|
| $p^2 = (ap)^2 = apap \Rightarrow p = apa$ | B1, 2 | For use of given properties to obtain AG. SR allow working from AG to obtain relevant properties |
**Part (ii)**
| $\left(p^2\right)^2 = p^4 = e \Rightarrow$ order $p^2 = 2$ | B1 | For correct order with no incorrect working seen |
|---|---|---|
| $\left(a^2\right)^2 = \left(p^2\right)^2 = e \Rightarrow$ order $a = 4$ | B1 | For correct order with no incorrect working seen |
|---|---|---|
| $(ap)^4 = a^4 = e \Rightarrow$ order $ap = 4$ | B1 | For correct order with no incorrect working seen |
|---|---|---|
| $\left(ap^2\right)^2 = ap^2ap^2 = ap \cdot a \cdot p = a^2$ | M1 | For relevant use of (i) or given properties |
|---|---|---|
| OR $ap^2 = a \cdot a^2 = a^3 \Rightarrow$ | A1, 5 | For correct order with no incorrect working seen |
|---|---|---|
| $\left(ap^2\right)^2 = a^6 = a^2$ | | |
| $\Rightarrow$ order $ap^2 = 4$ | | |
**Part (iii)**
**METHOD 1**
| $p^2 = a^2, ap^2 = a^3$ | M2 | For use of the given properties to simplify $p^2$ and $ap^2$ |
|---|---|---|
| $\Rightarrow [e, a, p^2, ap^2] = [e, a, a^2, a^3]$ | A1 | For obtaining $a^2$ and $a^3$ |
|---|---|---|
| which is a cyclic group | A1, 4 | For justifying that the set is a group |
**METHOD 2** (Cayley Table - Completion)
| | $e$ | $a$ | $p^2$ | $ap^2$ |
|---|---|---|---|---|
| $e$ | $e$ | $a$ | $p^2$ | $ap^2$ |
| $a$ | $a$ | $p^2$ | $ap^2$ | $e$ |
| $p^2$ | $p^2$ | $ap^2$ | $e$ | $a$ |
| $ap^2$ | $ap^2$ | $e$ | $a$ | $p^2$ |
| M1, A1 | For attempting closure with all 9 non-trivial products seen. For all 16 products correct |
|---|---|---|
| Completed table is a cyclic group | B2 | For justifying that the set is a group |
**METHOD 3** (Cayley Table - Identity, Inverses, Associativity)
| | $e$ | $a$ | $p^2$ | $ap^2$ |
|---|---|---|---|---|
| $e$ | $e$ | $a$ | $p^2$ | $ap^2$ |
| $a$ | $a$ | $p^2$ | $ap^2$ | $e$ |
| $p^2$ | $p^2$ | $ap^2$ | $e$ | $a$ |
| $ap^2$ | $ap^2$ | $e$ | $a$ | $p^2$ |
| M1, A1 | For attempting closure with all 9 non-trivial products seen. For all 16 products correct |
|---|---|---|
| Identity = $e$ | B1 | For stating identity |
|---|---|---|
| Inverses exist since EITHER: $e$ is in each row/column OR: $p^2$ is self-inverse; $a, ap^2$ form an inverse pair | B1 | For justifying inverses ($e^{-1} = e$ may be assumed) |
**Part (iv)**
**METHOD 1**
| e.g. $a \cdot ap = a^2p = p^3$ } $\Rightarrow$ not | M1 | For attempting to find a non-commutative pair of elements, at least one involving $a$ (may be embedded in a full or partial table) |
|---|---|---|
| $ap \cdot a = p$ | | |
| | M1, B1, A1, 4 | For simplifying elements both ways round. For a correct pair of non-commutative elements. For stating $Q$ non-commutative, with a clear argument |
**METHOD 2**
| Assume commutativity, so (eg) $ap = pa$ | M1 | For setting up proof by contradiction |
|---|---|---|
| (i) $\Rightarrow p = ap.a \Rightarrow pa.a = pa^2 = pp^2 = p^3$ | M1 | For using (i) and/or given properties |
|---|---|---|
| But $p$ and $p^3$ are distinct | B1 | For obtaining and stating a contradiction |
|---|---|---|
| $\Rightarrow Q$ is non-commutative | A1 | For stating $Q$ non-commutative, with a clear argument |
---8 A multiplicative group $Q$ of order 8 has elements $\left\{ e , p , p ^ { 2 } , p ^ { 3 } , a , a p , a p ^ { 2 } , a p ^ { 3 } \right\}$, where $e$ is the identity. The elements have the properties $p ^ { 4 } = e$ and $a ^ { 2 } = p ^ { 2 } = ( a p ) ^ { 2 }$.\\
(i) Prove that $a = p a p$ and that $p = a p a$.\\
(ii) Find the order of each of the elements $p ^ { 2 } , a , a p , a p ^ { 2 }$.\\
(iii) Prove that $\left\{ e , a , p ^ { 2 } , a p ^ { 2 } \right\}$ is a subgroup of $Q$.\\
(iv) Determine whether $Q$ is a commutative group.
\hfill \mbox{\textit{OCR FP3 2009 Q8 [15]}}