| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Subgroups and cosets |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring understanding of group axioms, orders of elements, and subgroup structure. Part (i) involves algebraic manipulation using given relations, (ii) requires proving orders by calculation, and (iii) demands identification of Klein-4 subgroups and proving non-cyclicity. While systematic, it requires solid abstract algebra knowledge beyond standard A-level and multi-step reasoning through unfamiliar structures. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(wa^2 = waa = a^3wa = a^3a^3w\) | M1 | Use \(wa = a^3w\) to simplify |
| \(= a^4a^2w = ea^2w\) | B1 | Use \(a^4 = e\) (oe) in either proof |
| \(= a^2w\) | A1 | Complete argument AG |
| Either result \(\Rightarrow wa^3 = a^3wa^2\) | M1 | |
| \(= a^3a^2w\) | M1 | |
| \(= eaw = aw\) | A1 | AG |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((aw)^2 = (aw)(aw) = awwa^3 = aea^3 = a^4 = e\) so order 2 | M1 | For squaring any of elements |
| M1 | For attempt to simplify to \(e\) | |
| \((a^2w)(a^2w) = a^2wwa^2 = a^2ea^2 = a^4 = e\) so order 2 | A1 | For at least two squared elements shown equal to \(e\) |
| \((a^3w)(a^3w) = a^3wwa = a^3ea = a^4 = e\) so order 2 | A1 | For complete argument |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\{e, a^2, w, a^2w\}\) | B1 | Condone equivalents |
| \(\{e, a^2, aw, a^3w\}\) | B1 | |
| \(a^2, w, aw, a^2w, a^3w\) all of order 2 | M1 | Consider orders; or considers form \(\{e, x, y, xy\}\) where \(x\), \(y\) order 2 |
| so not cyclic as no element of order 4 in either | A1 | Dep on both groups correct; condone 'no generator' or 'Klein (V) group' in place of 'no element of order 4' |
| [4] |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $wa^2 = waa = a^3wa = a^3a^3w$ | M1 | Use $wa = a^3w$ to simplify |
| $= a^4a^2w = ea^2w$ | B1 | Use $a^4 = e$ (oe) in either proof |
| $= a^2w$ | A1 | Complete argument AG |
| Either result $\Rightarrow wa^3 = a^3wa^2$ | M1 | |
| $= a^3a^2w$ | M1 | |
| $= eaw = aw$ | A1 | AG |
| **[6]** | | |
---
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(aw)^2 = (aw)(aw) = awwa^3 = aea^3 = a^4 = e$ so order 2 | M1 | For squaring any of elements |
| | M1 | For attempt to simplify to $e$ |
| $(a^2w)(a^2w) = a^2wwa^2 = a^2ea^2 = a^4 = e$ so order 2 | A1 | For at least two squared elements shown equal to $e$ |
| $(a^3w)(a^3w) = a^3wwa = a^3ea = a^4 = e$ so order 2 | A1 | For complete argument |
| **[4]** | | |
---
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\{e, a^2, w, a^2w\}$ | B1 | Condone equivalents |
| $\{e, a^2, aw, a^3w\}$ | B1 | |
| $a^2, w, aw, a^2w, a^3w$ all of order 2 | M1 | Consider orders; or considers form $\{e, x, y, xy\}$ where $x$, $y$ order 2 |
| so not cyclic as no element of order 4 in either | A1 | Dep on both groups correct; condone 'no generator' or 'Klein (V) group' in place of 'no element of order 4' |
| **[4]** | | |8 A multiplicative group $H$ has the elements $\left\{ e , a , a ^ { 2 } , a ^ { 3 } , w , a w , a ^ { 2 } w , a ^ { 3 } w \right\}$ where $e$ is the identity, elements $a$ and $w$ have orders 4 and 2 respectively and $w a = a ^ { 3 } w$.\\
(i) Show that $w a ^ { 2 } = a ^ { 2 } w$ and also that $w a ^ { 3 } = a w$.\\
(ii) Hence show that each of $a w , a ^ { 2 } w$ and $a ^ { 3 } w$ has order 2 .\\
(iii) Find two non-cyclic subgroups of $H$ of order 4, and show that they are not cyclic.
\hfill \mbox{\textit{OCR FP3 2013 Q8 [14]}}