| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Isomorphism between groups |
| Difficulty | Challenging +1.8 This is a substantial Further Maths group theory question requiring multiple techniques: verifying group axioms, understanding isomorphism theory (including the key result about groups of prime order), constructing explicit isomorphisms, working with direct products, and counting subgroups. While each individual part is methodical, the question demands sustained abstract reasoning across 8 parts, making it significantly harder than standard A-level content but still within reach for well-prepared FM students. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups8.03h Generators: of cyclic and non-cyclic groups8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | 3 | 4 |
| 1 | 1 | 3 |
| 3 | 3 | 9 |
| 4 | 4 | 1 |
| 5 | 5 | 4 |
| 9 | 9 | 5 |
| Composition table shows closure | B1 | |
| Identity is 1 | B1 | Dependent on B2 for table |
| Answer | Marks | Guidance |
|---|---|---|
| Element | 1 | 3 |
| Inverse | 1 | 4 |
| So every element has an inverse | (6 marks) |
| Answer | Marks |
|---|---|
| Since 5 is prime, a group of order 5 must be cyclic | B1 B1 |
| Two cyclic groups of the same order must be isomorphic | B1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H\) | 1 | \(e^{\frac{2\pi i}{5}}\) |
| \(G\) | 1 | 3 |
| or | 1 | 4 |
| or | 1 | 5 |
| or | 1 | 9 |
| Answer | Marks |
|---|---|
| Identity is \((1, 1)\) | B1 |
| Inverse of \((9, 3)\) is \((5, 4)\) | B1 (2 marks) |
| Answer | Marks |
|---|---|
| \((x, y)^5 = (x^5, y^5)\) | M1 |
| Since \(G\) has order 5, \(x^5 = 1\) and \(y^5 = 1\) | M1 A1 (ag) |
| Hence \((x, y)^5 = (1, 1)\) | A1 (ag) (3 marks) |
| Answer | Marks |
|---|---|
| Order of \((x, y)\) is a factor of 5 (so must be 1 or 5) | M1 |
| Only identity \((1, 1)\) can have order 1 | B1 A1 (ag) |
| Hence all other elements have order 5 | A1 (ag) (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{(1, 1), (9, 3), (4, 9), (3, 5), (5, 4)\}\) | B2 | Give B1 if for 5 elements including \((1, 1)\), \((9, 3)\), \((5, 4)\) (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| An element of order 5 generates a subgroup, and so can be in only one subgroup of order 5 | M1 | Or for 24 ÷ 4 or Or listing at least 2 other subgroups |
| Number is \(24 \div 4 = 6\) | A1 | Solving to obtain a value of \(p\) (2 marks); Give B1 for unsupported answer 6 |
### 4(i)
**Composition table:**
| | 1 | 3 | 4 | 5 | 9 |
|---|---|---|---|---|---|
| 1 | 1 | 3 | 4 | 5 | 9 |
| 3 | 3 | 9 | 1 | 4 | 5 |
| 4 | 4 | 1 | 5 | 9 | 3 |
| 5 | 5 | 4 | 9 | 3 | 1 |
| 9 | 9 | 5 | 3 | 1 | 4 | B2 | Give B1 if not more than 4 errors
Composition table shows closure | B1 |
Identity is 1 | B1 | Dependent on B2 for table
**Inverse table:**
| Element | 1 | 3 | 4 | 5 | 9 |
|---------|---|---|---|---|---|
| Inverse | 1 | 4 | 3 | 9 | 5 | B2 | Give B1 for 3 correct
So every element has an inverse | (6 marks)
### 4(ii)
Since 5 is prime, a group of order 5 must be cyclic | B1 B1 |
Two cyclic groups of the same order must be isomorphic | B1 (3 marks)
### 4(iii)
| $H$ | 1 | $e^{\frac{2\pi i}{5}}$ | $e^{\frac{4\pi i}{5}}$ | $e^{\frac{6\pi i}{5}}$ | $e^{\frac{8\pi i}{5}}$ |
|-----|---|---|---|---|---|
| $G$ | 1 | 3 | 9 | 5 | 4 | B1 | For $1 \leftrightarrow 1$
or | 1 | 4 | 5 | 9 | 3 | B2 | For non-identity elements (3 marks)
or | 1 | 5 | 3 | 4 | 9 |
or | 1 | 9 | 4 | 3 | 5 |
### 4(iv)
Identity is $(1, 1)$ | B1 |
Inverse of $(9, 3)$ is $(5, 4)$ | B1 (2 marks)
### 4(v)
$(x, y)^5 = (x^5, y^5)$ | M1 |
Since $G$ has order 5, $x^5 = 1$ and $y^5 = 1$ | M1 A1 (ag) |
Hence $(x, y)^5 = (1, 1)$ | A1 (ag) (3 marks)
### 4(vi)
Order of $(x, y)$ is a factor of 5 (so must be 1 or 5) | M1 |
Only identity $(1, 1)$ can have order 1 | B1 A1 (ag) |
Hence all other elements have order 5 | A1 (ag) (3 marks)
### 4(vii)
$\{(1, 1), (9, 3), (4, 9), (3, 5), (5, 4)\}$ | B2 | Give B1 if for 5 elements including $(1, 1)$, $(9, 3)$, $(5, 4)$ (2 marks)
### 4(viii)
An element of order 5 generates a subgroup, and so can be in only one subgroup of order 5 | M1 | Or for 24 ÷ 4 or Or listing at least 2 other subgroups
Number is $24 \div 4 = 6$ | A1 | Solving to obtain a value of $p$ (2 marks); Give B1 for unsupported answer 64 (i) Show that the set $G = \{ 1,3,4,5,9 \}$, under the binary operation of multiplication modulo 11 , is a group. You may assume associativity.\\
(ii) Explain why any two groups of order 5 must be isomorphic to each other.
The set $H = \left\{ 1 , \mathrm { e } ^ { \frac { 2 } { 5 } \pi \mathrm { j } } , \mathrm { e } ^ { \frac { 4 } { 5 } \pi \mathrm { j } } , \mathrm { e } ^ { \frac { 6 } { 5 } \pi \mathrm { j } } , \mathrm { e } ^ { \frac { 8 } { 5 } \pi \mathrm { j } } \right\}$ is a group under the binary operation of multiplication of complex numbers.\\
(iii) Specify an isomorphism between the groups $G$ and $H$.
The set $K$ consists of the 25 ordered pairs $( x , y )$, where $x$ and $y$ are elements of $G$. The set $K$ is a group under the binary operation defined by
$$\left( x _ { 1 } , y _ { 1 } \right) \left( x _ { 2 } , y _ { 2 } \right) = \left( x _ { 1 } x _ { 2 } , y _ { 1 } y _ { 2 } \right)$$
where the multiplications are carried out modulo 11 ; for example, $( 3,5 ) ( 4,4 ) = ( 1,9 )$.\\
(iv) Write down the identity element of $K$, and find the inverse of the element $( 9,3 )$.\\
(v) Explain why $( x , y ) ^ { 5 } = ( 1,1 )$ for every element $( x , y )$ in $K$.\\
(vi) Deduce that all the elements of $K$, except for one, have order 5. State which is the exceptional element.\\
(vii) A subgroup of $K$ has order 5 and contains the element (9, 3). List the elements of this subgroup.\\
(viii) Determine how many subgroups of $K$ there are with order 5 .
\hfill \mbox{\textit{OCR MEI FP3 2011 Q4 [24]}}