| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Prove group-theoretic identities |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring proof by contradiction, computation of element orders in a modular arithmetic group, identification of subgroups, and isomorphism analysis. While the individual techniques are standard for FP2, the combination of abstract proof, systematic enumeration, and structural comparison requires solid understanding of group theory concepts beyond typical A-level. The proof in part (i) is elegant but non-trivial; parts (ii)(a-b) involve substantial computation; part (c) requires recognizing cyclic group structure. This is appropriately challenging for Further Maths but not exceptionally difficult within that context. |
| Spec | 8.03a Binary operations: and their properties on given sets8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assume \(ab = ba\); as \(a^2b = ba\) then \(ab = a^2b\) | M1 | Proof begins with assumption \(ab = ba\), deduces \(ab = a^2b\) |
| So \(a^{-1}abb^{-1} = a^{-1}a^2bb^{-1}\) | M1 | Correct proof with working shown |
| So \(e = a\) | A1 | Concludes assumption implies \(e = a\) |
| Contradiction as \(e\) and \(a\) are distinct, so \(ab \neq ba\) | A1 | Clearly explains contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 2 has order 4 and 4 has order 2 | M1 | Obtains two correct orders |
| 7, 8 and 13 have order 4 | A1 | Finds another three correctly |
| 11 and 14 have order 2, and 1 has order 1 | A1 | Finds final three so all eight are correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Finds subgroup \(\{1, 2, 4, 8\}\) or \(\{1, 7, 4, 13\}\) | M1 | Finds one cyclic subgroup |
| Finds both, refers to them as cyclic groups, or gives generators 2 and 7 | A1 | Both subgroups identified as cyclic |
| Finds \(\{1, 4, 11, 14\}\) | B1 | Non-cyclic group found |
| States each element has order 2 or refers to it as Klein Group | B1 | Correct description of non-cyclic group |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(J\) has an element of order 8, \(H\) does not; or \(J\) is cyclic, \(H\) is not; or other valid reason | M1 | Clearly explains how \(J\) differs from \(H\) |
| They are not isomorphic | A1 | Correct deduction |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume $ab = ba$; as $a^2b = ba$ then $ab = a^2b$ | M1 | Proof begins with assumption $ab = ba$, deduces $ab = a^2b$ |
| So $a^{-1}abb^{-1} = a^{-1}a^2bb^{-1}$ | M1 | Correct proof with working shown |
| So $e = a$ | A1 | Concludes assumption implies $e = a$ |
| Contradiction as $e$ and $a$ are distinct, so $ab \neq ba$ | A1 | Clearly explains contradiction |
### Part (ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 2 has order 4 and 4 has order 2 | M1 | Obtains two correct orders |
| 7, 8 and 13 have order 4 | A1 | Finds another three correctly |
| 11 and 14 have order 2, and 1 has order 1 | A1 | Finds final three so all eight are correct |
### Part (ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds subgroup $\{1, 2, 4, 8\}$ or $\{1, 7, 4, 13\}$ | M1 | Finds one cyclic subgroup |
| Finds both, refers to them as cyclic groups, or gives generators 2 and 7 | A1 | Both subgroups identified as cyclic |
| Finds $\{1, 4, 11, 14\}$ | B1 | Non-cyclic group found |
| States each element has order 2 or refers to it as Klein Group | B1 | Correct description of non-cyclic group |
### Part (ii)(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $J$ has an element of order 8, $H$ does not; or $J$ is cyclic, $H$ is not; or other valid reason | M1 | Clearly explains how $J$ differs from $H$ |
| They are not isomorphic | A1 | Correct deduction |
---\begin{enumerate}
\item (i) A group $G$ contains distinct elements $a , b$ and $e$ where $e$ is the identity element and the group operation is multiplication.
\end{enumerate}
Given $a ^ { 2 } b = b a$, prove $a b \neq b a$\\
(ii) The set $H = \{ 1,2,4,7,8,11,13,14 \}$ forms a group under the operation of multiplication modulo 15\\
(a) Find the order of each element of $H$.\\
(b) Find three subgroups of $H$ each of order 4, and describe each of these subgroups.
The elements of another group $J$ are the matrices $\left( \begin{array} { c c } \cos \left( \frac { k \pi } { 4 } \right) & \sin \left( \frac { k \pi } { 4 } \right) \\ - \sin \left( \frac { k \pi } { 4 } \right) & \cos \left( \frac { k \pi } { 4 } \right) \end{array} \right)$\\
where $k = 1,2,3,4,5,6,7,8$ and the group operation is matrix multiplication.\\
(c) Determine whether $H$ and $J$ are isomorphic, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel FP2 Q4 [13]}}