| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Order of elements |
| Difficulty | Challenging +1.8 This FP3 group theory question requires understanding of element orders, Lagrange's theorem, and subgroup structure. Part (i) involves systematic calculation using order properties (4 marks of routine application). Part (ii) requires constructing element-order tables for Klein-4 and S_3, then using these to deduce subgroup impossibility—this demands deeper conceptual reasoning about subgroup structure rather than just computation. While systematic, it's non-trivial problem-solving in abstract algebra, a topic inherently harder than standard A-level content. |
| 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 |
| Order of element | 1 | 2 | 3 | 4 | 6 |
| Number of elements | 1 | 1 | 2 | 6 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a^4 = r^6 = e \Rightarrow a\) has order 4, \(a^2\) has order 2 | M1 | For considering powers of \(a\) |
| \((a^3)^4 = a^{12} = e \Rightarrow a^3\) has order 4 | A1 | For order of any one of \(a, a^2, a^3\) correct |
| A1 | For all correct | |
| \((r^2)^3 = e \Rightarrow r^2\) has order 3 | B1 4 | For order of \(r^2\) correct |
| Answer | Marks |
|---|---|
| M1 | For top line in either table. Allow inclusion of 4 and 6 respectively (and other orders if 0 appears below) |
| Answer | Marks | Guidance |
|---|---|---|
| Order of element | 1 | 2 |
| Number of elements | 1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Order of element | 1 | 2 |
| Number of elements | 1 | 3 |
| A1 | For order 4 table | |
| A1 | For order 6 table | |
| B1 | For stating that only \(G\) and \(H\) need be considered AEF | |
| B1 5 | For argument completed by elements of order 2 AG SR Allow equivalent arguments for B1 B1 |
## (i)
$a^4 = r^6 = e \Rightarrow a$ has order 4, $a^2$ has order 2 | M1 | For considering powers of $a$
$(a^3)^4 = a^{12} = e \Rightarrow a^3$ has order 4 | A1 | For order of any one of $a, a^2, a^3$ correct
| A1 | For all correct
$(r^2)^3 = e \Rightarrow r^2$ has order 3 | B1 4 | For order of $r^2$ correct
## (ii)
| M1 | For top line in either table. Allow inclusion of 4 and 6 respectively (and other orders if 0 appears below)
**G order 4**
| Order of element | 1 | 2 | (4) |
|---|---|---|---|
| Number of elements | 1 | 1 | (0) |
**H order 6**
| Order of element | 1 | 2 | 3 | (6) |
|---|---|---|---|---|
| Number of elements | 1 | 3 | 2 | (0) |
| A1 | For order 4 table
| A1 | For order 6 table
| B1 | For stating that only $G$ and $H$ need be considered **AEF**
| B1 5 | For argument completed by elements of order 2 **AG** SR Allow equivalent arguments for B1 B1
---$Q$ is a multiplicative group of order 12.
\begin{enumerate}[label=(\roman*)]
\item Two elements of $Q$ are $a$ and $r$. It is given that $r$ has order 6 and that $a^2 = r^3$. Find the orders of the elements $a$, $a^2$, $a^3$ and $r^2$. [4]
\end{enumerate}
The table below shows the number of elements of $Q$ with each possible order.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Order of element & 1 & 2 & 3 & 4 & 6 \\
\hline
Number of elements & 1 & 1 & 2 & 6 & 2 \\
\hline
\end{tabular}
\end{center}
$G$ and $H$ are the non-cyclic groups of order 4 and 6 respectively.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Construct two tables, similar to the one above, to show the number of elements with each possible order for the groups $G$ and $H$. Hence explain why there are no non-cyclic proper subgroups of $Q$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q6 [9]}}