| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Order of elements and cyclic structure |
| Difficulty | Standard +0.8 This is a Further Maths group theory question requiring understanding of element orders, subgroups, cyclic groups, and isomorphism to complex numbers. While systematic, it demands multiple abstract concepts beyond standard A-level and involves non-trivial reasoning about group structure, placing it moderately above average difficulty. |
| Spec | 8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups8.03g Cyclic groups: meaning of the term8.03h Generators: of cyclic and non-cyclic groups |
| \(*\) | \(a\) | \(b\) | \(c\) | \(d\) |
| \(a\) | \(d\) | \(a\) | \(b\) | \(c\) |
| \(b\) | \(a\) | \(b\) | \(c\) | \(d\) |
| \(c\) | \(b\) | \(c\) | \(d\) | \(a\) |
| \(d\) | \(c\) | \(d\) | \(a\) | \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(b\) is the identity and so has order 1 | B1 | For identifying \(b\) as the identity element |
| \(d * d = b\), so \(d\) has order 2 | B1 | For stating the order of \(d\) is 2 |
| \(a * a = c * c = d\), so \(a\) and \(c\) each have order 4 | B1 | For both orders stated |
| 3 | ||
| (ii) \(\{b, d\}\) | B1 | For stating this subgroup |
| 1 | ||
| (iii) \(G\) is cyclic because it has an element of order 4 | B1 | For correct answer with justification |
| 1 | ||
| (iv) \(b = 1, d = -1, a = i, c = -i\) (or vice versa for \(a, c\)) | B1 | For all four correct values |
| 1 | ||
| 6 |
**(i)** $b$ is the identity and so has order 1 | B1 | For identifying $b$ as the identity element
$d * d = b$, so $d$ has order 2 | B1 | For stating the order of $d$ is 2
$a * a = c * c = d$, so $a$ and $c$ each have order 4 | B1 | For both orders stated
| **3** |
**(ii)** $\{b, d\}$ | B1 | For stating this subgroup
| **1** |
**(iii)** $G$ is cyclic because it has an element of order 4 | B1 | For correct answer with justification
| **1** |
**(iv)** $b = 1, d = -1, a = i, c = -i$ (or vice versa for $a, c$) | B1 | For all four correct values
| **1** |
| **6** |2 The set $S = \{ a , b , c , d \}$ under the binary operation * forms a group $G$ of order 4 with the following operation table.
\begin{center}
\begin{tabular}{ l | l l l l }
$*$ & $a$ & $b$ & $c$ & $d$ \\
\hline
$a$ & $d$ & $a$ & $b$ & $c$ \\
$b$ & $a$ & $b$ & $c$ & $d$ \\
$c$ & $b$ & $c$ & $d$ & $a$ \\
$d$ & $c$ & $d$ & $a$ & $b$ \\
\end{tabular}
\end{center}
(i) Find the order of each element of $G$.\\
(ii) Write down a proper subgroup of $G$.\\
(iii) Is the group $G$ cyclic? Give a reason for your answer.\\
(iv) State suitable values for each of $a , b , c$ and $d$ in the case where the operation $*$ is multiplication of complex numbers.
\hfill \mbox{\textit{OCR FP3 Q2 [6]}}