| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Isomorphism between groups |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring understanding of subgroup orders (Lagrange's theorem), cyclic subgroups, identity/order concepts, and isomorphisms. While the individual parts are mostly computational applications of definitions, part (f) requires conceptual understanding of why a bijection isn't necessarily an isomorphism. The modular arithmetic context and multi-part structure with proof elements places this significantly above average A-level difficulty. |
| Spec | 8.03f Subgroups: definition and tests for proper subgroups8.03h Generators: of cyclic and non-cyclic groups8.03k Lagrange's theorem: order of subgroup divides order of group8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | 2, 3, 4 and 6 |
| because these are the factors of 12 (by Lagrange’s Theorem) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 2.4 | Ignore 1 or 12 if they appear | |
| (b) | {3, 9, 27} | B1 |
| [1] | 1.1 | |
| (c) | 27 | B1 |
| [1] | 1.1 | |
| (d) | 18, 182 = 324 =12 mod (39), (183 = 21) … | |
| 184 = 122 = 27 (mod 39) so order 4 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.1 | For method of searching by finding powers |
| (e) | 3 ≡ 3, 42, 81, … |
| Answer | Marks |
|---|---|
| Second square-root of 3 is −9 ≡ 30 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 3.2a | Method for searching for squares ≡ 3 (mod 39) |
| Answer | Marks | Guidance |
|---|---|---|
| (f) | Identity 27 in G does not map to identity 1 in H | B1 |
| [1] | 2.4 | Or any clear equivalent statement |
Question 6:
6 | (a) | 2, 3, 4 and 6
because these are the factors of 12 (by Lagrange’s Theorem) | B1
B1
[2] | 1.1
2.4 | Ignore 1 or 12 if they appear
(b) | {3, 9, 27} | B1
[1] | 1.1
(c) | 27 | B1
[1] | 1.1
(d) | 18, 182 = 324 =12 mod (39), (183 = 21) …
184 = 122 = 27 (mod 39) so order 4 | M1
A1
[2] | 1.1
2.1 | For method of searching by finding powers
(e) | 3 ≡ 3, 42, 81, …
First square-root of 3 is 9
Second square-root of 3 is −9 ≡ 30 | M1
A1
A1
[3] | 3.1a
2.2a
3.2a | Method for searching for squares ≡ 3 (mod 39)
Cannot just state −9
(f) | Identity 27 in G does not map to identity 1 in H | B1
[1] | 2.4 | Or any clear equivalent statement6 The group $G$ consists of the set $\{ 3,6,9,12,15,18,21,24,27,30,33,36 \}$ under $\times _ { 39 }$, the operation of multiplication modulo 39.
\begin{enumerate}[label=(\alph*)]
\item List the possible orders of proper subgroups of $G$, justifying your answer.
\item List the elements of the subset of $G$ generated by the element 3 .
\item State the identity element of $G$.
\item Determine the order of the element 18 .
\item Find the two elements $g _ { 1 }$ and $g _ { 2 }$ in $G$ which satisfy $g \times { } _ { 39 } g = 3$.
The group $H$ consists of the set $\{ 1,2,3,4,5,6,7,8,9,10,11,12 \}$ under $\times _ { 13 }$, the operation of multiplication modulo 13. You are given that $G$ is isomorphic to $H$.
A student states that $G$ is isomorphic to $H$ because each element $3 x$ in $G$ maps directly to the element $x$ in $H$ (for $x = 1,2,3,4,5,6,7,8,9,10,11,12$ ).
\item Explain why this student is incorrect.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2020 Q6 [10]}}