| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Order of elements and cyclic structure |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring proof techniques and understanding of element orders in groups. Part (i) needs systematic verification that (ab)^6 = e and no smaller power works, using commutativity and order properties. Part (ii) requires recognizing that an element of order 18 exists (constructible from the given elements), demonstrating G is cyclic. While the concepts are advanced, the proof steps follow standard group theory arguments that well-prepared FM students should know, making it challenging but not exceptional. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03g Cyclic groups: meaning of the term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((ab)^6 = ababab = a^6b^6\) as commutative | M1 | Must give reason. Some demonstration that they understand commutativity |
| \(= (a^2)^3(b^3)^2 = e^3e^2 = e\) | A1 | Using orders of \(a\) and \(b\) |
| So \(ab\) has order 1, 2, 3, or 6 | ||
| \((b\neq a \Rightarrow ab\neq a^2 \Rightarrow ab\neq e\) so \(ab\) not order 1) | Condone absence of this line | |
| \((ab)^2 = a^2b^2 = eb^2 = b^2\) and \(b\) not order 2, so \(ab\) not order 2 | M1 | Consider other cases. Insufficient to merely have simplified all \((ab)^n\) |
| \((ab)^3 = a^3b^3 = aa^2e = aee = a \neq e\), so \(ab\) not order 3 | ||
| (So must be order 6) | A1 | AG Complete argument |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(ac\) has order 18 | B1 | Or \(abc\) or generator |
| 18 is LCM of 2 and 9, (so we can use a similar argument to part (i)) | M1 | or explicit consideration of other cases |
| So as \(G\) has an element of order 18 it must be cyclic. | A1 | AG Complete argument |
| [3] |
# Question 7:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(ab)^6 = ababab = a^6b^6$ as commutative | M1 | Must give reason. Some demonstration that they understand commutativity |
| $= (a^2)^3(b^3)^2 = e^3e^2 = e$ | A1 | Using orders of $a$ and $b$ |
| So $ab$ has order 1, 2, 3, or 6 | | |
| $(b\neq a \Rightarrow ab\neq a^2 \Rightarrow ab\neq e$ so $ab$ not order 1) | | Condone absence of this line |
| $(ab)^2 = a^2b^2 = eb^2 = b^2$ and $b$ not order 2, so $ab$ not order 2 | M1 | Consider other cases. Insufficient to merely have simplified **all** $(ab)^n$ |
| $(ab)^3 = a^3b^3 = aa^2e = aee = a \neq e$, so $ab$ not order 3 | | |
| (So must be order 6) | A1 | AG Complete argument |
| | **[4]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $ac$ has order 18 | B1 | Or $abc$ or generator |
| 18 is LCM of 2 and 9, (so we can use a similar argument to part (i)) | M1 | or explicit consideration of other cases |
| So as $G$ has an element of order 18 it must be cyclic. | A1 | AG Complete argument |
| | **[3]** | |
---7 A commutative group $G$ has order 18. The elements $a , b$ and $c$ have orders 2, 3 and 9 respectively.\\
(i) Prove that $a b$ has order 6 .\\
(ii) Show that $G$ is cyclic.
\hfill \mbox{\textit{OCR FP3 2013 Q7 [7]}}