| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| 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 Lagrange's theorem, element orders, and subgroup structure. While the concepts are advanced, the question guides students through systematic exploration of a specific group of order 9, making it more accessible than open-ended abstract algebra proofs. The multi-part structure and requirement to justify orders elevates it above routine exercises. |
| Spec | 8.03e 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (a) \(e, p, p^2\) | B1 | For correct elements |
| (b) \(e, q, q^2\) | B1 2 | For correct elements; SR if answers to (i) and (iv) are reversed, full credit may be earned for both parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p^3=q^3=e\Rightarrow(pq)^3=p^3q^3=e\) | M1 | For finding \((pq)^3\) or \((pq^2)^3\) |
| \(\Rightarrow\) order 3 | A1 | For correct order |
| \((pq^2)^3=p^3q^6=p^3(q^3)^2=e\Rightarrow\) order 3 | A1 3 | For correct order; SR for answer(s) only allow B1 for either or both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3\) | B1 1 | For correct order and no others |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e\), \(pq\), \(p^2q^2\) OR \(e\), \(pq\), \((pq)^2\) | B1 | For stating \(e\) and either \(pq\) or \(p^2q^2\) |
| B1 | For all 3 elements and no more | |
| \(e\), \(pq^2\), \(p^2q^4\) OR \(e\), \(pq^2\), \((pq^2)^2\) | B1 | For stating \(e\) and either \(pq^2\) or \(p^2q\) |
| OR \(e\), \(p^2q\), \((p^2q)^2\) | B1 4 | For all 3 elements and no more |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(a)** $e, p, p^2$ | B1 | For correct elements |
| **(b)** $e, q, q^2$ | B1 **2** | For correct elements; SR if answers to (i) and (iv) are reversed, full credit may be earned for both parts |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p^3=q^3=e\Rightarrow(pq)^3=p^3q^3=e$ | M1 | For finding $(pq)^3$ or $(pq^2)^3$ |
| $\Rightarrow$ order 3 | A1 | For correct order |
| $(pq^2)^3=p^3q^6=p^3(q^3)^2=e\Rightarrow$ order 3 | A1 **3** | For correct order; SR for answer(s) only allow B1 for either or both |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3$ | B1 **1** | For correct order and no others |
### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e$, $pq$, $p^2q^2$ OR $e$, $pq$, $(pq)^2$ | B1 | For stating $e$ and either $pq$ or $p^2q^2$ |
| | B1 | For all 3 elements and no more |
| $e$, $pq^2$, $p^2q^4$ OR $e$, $pq^2$, $(pq^2)^2$ | B1 | For stating $e$ and either $pq^2$ or $p^2q$ |
| OR $e$, $p^2q$, $(p^2q)^2$ | B1 **4** | For all 3 elements and no more |
---5 A multiplicative group $G$ of order 9 has distinct elements $p$ and $q$, both of which have order 3 . The group is commutative, the identity element is $e$, and it is given that $q \neq p ^ { 2 }$.\\
(i) Write down the elements of a proper subgroup of $G$
\begin{enumerate}[label=(\alph*)]
\item which does not contain $q$,
\item which does not contain $p$.\\
(ii) Find the order of each of the elements $p q$ and $p q ^ { 2 }$, justifying your answers.\\
(iii) State the possible order(s) of proper subgroups of $G$.\\
(iv) Find two proper subgroups of $G$ which are distinct from those in part (i), simplifying the elements.
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2007 Q5 [10]}}