| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Challenging +1.2 This is a structured group theory question with clear scaffolding. Part (i) is algebraic manipulation to verify closure, part (ii) is computational (constructing a 5×5 table), part (iii) applies standard group axioms with associativity given, and part (iv) requires identifying generators from the table. While group theory is a Further Maths topic making it inherently harder than Core content, the question is methodical with no novel insights required—students follow a standard verification procedure. The algebraic proof in (i) is straightforward expansion and factoring. |
| Spec | 8.02e Finite (modular) arithmetic: integers modulo n8.03c Group definition: recall and use, show structure is/isn't a group8.03d Latin square property: for group tables8.03h Generators: of cyclic and non-cyclic groups |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) | (cid:32)100ab(cid:14)60(a(cid:14)b)(cid:14)36 |
| Answer | Marks |
|---|---|
| even + even + odd | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | Correctly multiplied out given |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (ii) | 16 36 56 76 96 |
| Answer | Marks |
|---|---|
| 96 36 56 76 96 16 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | For most entries correct ((cid:100)2 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (iii) | Closure noted – no new elements in the table |
| Answer | Marks |
|---|---|
| Inverse-pairs 16, 36 and 56, 96 noted | B1 |
| Answer | Marks |
|---|---|
| [3] | 2.5 |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | Or from (i) | |
| 4 | (iv) | 16, 36, 56, 96 are all possible generators |
| [1] | 2.2a | |
| 16 | 36 | 56 |
Question 4:
4 | (i) | (cid:32)100ab(cid:14)60(a(cid:14)b)(cid:14)36
(cid:32)10(cid:11)10ab(cid:14)6(a(cid:14)b)(cid:14)3(cid:12)(cid:14)6
and n(cid:32)10ab(cid:14)6(cid:11)a(cid:14)b(cid:12)(cid:14)3 is odd since it is
even + even + odd | M1
A1
E1
[3] | 1.1
3.1a
2.4 | Correctly multiplied out given
expression
Correctly identified “tens” term
Proper explanation that n is odd
4 | (ii) | 16 36 56 76 96
16 56 76 96 16 36
36 76 96 16 36 56
56 96 16 36 56 76
76 16 36 56 76 96
96 36 56 76 96 16 | M1
A1
[2] | 1.1
1.1 | For most entries correct ((cid:100)2
errors)
For all entries correct
4 | (iii) | Closure noted – no new elements in the table
(Associativity given)
76 is the identity
Inverse-pairs 16, 36 and 56, 96 noted | B1
B1
B1
[3] | 2.5
1.2
1.1 | Or from (i)
4 | (iv) | 16, 36, 56, 96 are all possible generators | B1
[1] | 2.2a
16 | 36 | 56 | 76 | 964 Let $S$ be the set $\{ 16,36,56,76,96 \}$ and $\times _ { H }$ the operation of multiplication modulo 100 .\\
(i) Given that $a$ and $b$ are odd positive integers, show that $( 10 a + 6 ) ( 10 b + 6 )$ can also be written in the form $10 n + 6$ for some odd positive integer $n$.\\
(ii) Construct the Cayley table for $\left( S , \times _ { H } \right)$\\
(iii) Show that $\left( S , \times _ { H } \right)$ is a group.\\[0pt]
[You may use the result that $\times _ { H }$ is associative on $S$.]\\
(iv) Write down all generators of $\left( S , \times _ { H } \right)$.
\hfill \mbox{\textit{OCR Further Additional Pure AS Q4 [9]}}