| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2017 |
| Session | December |
| Marks | 11 |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Challenging +1.2 This is a Further Maths group theory question requiring verification of group axioms and construction of a Cayley table. While it involves abstract algebra (inherently harder than core A-level), the tasks are methodical: proving associativity through algebraic manipulation, completing a small Cayley table by calculation, verifying closure/identity/inverses from the table, and identifying subgroups. These are standard textbook exercises in group theory rather than requiring novel insight, placing it moderately above average difficulty. |
| Spec | 8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group8.03f Subgroups: definition and tests for proper subgroups |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((a \diamond b) \diamond c = (a + b - ab) \diamond c = (a + b - ab) + c - (a + b - ab)c = a + b + c - (ab + bc + ca) + abc\); \(a \diamond (b \diamond c) = a \diamond (b + c - bc) = a + (b + c - bc) - a(b + c - bc) = a + b + c - (ab + bc + ca) + abc\) | M1 A1 A1 [3] | 1.2 1.1 1.1 |
| (ii) | (A, ○) | 0 |
| 0 | 0 | 2 |
| 2 | 2 | 0 |
| 3 | 3 | 6 |
| 4 | 4 | 5 |
| 5 | 5 | 4 |
| 6 | 6 | 3 |
| (iii) Closure follows from table (no new elements); (Associativity follows from part (i)); 0 is the identity; 2 is self-inverse; (3, 5) and (4, 6) are inverse-pairs, so every (non-identity) element has an inverse in the set | B1 B1 B1 [3] | 2.2a 2.2a 2.1 |
| (iv) {0}, {0, 2}, {0, 4, 6}, A | B1 B1 [2] | 1.1 2.5 |
**(i)** $(a \diamond b) \diamond c = (a + b - ab) \diamond c = (a + b - ab) + c - (a + b - ab)c = a + b + c - (ab + bc + ca) + abc$; $a \diamond (b \diamond c) = a \diamond (b + c - bc) = a + (b + c - bc) - a(b + c - bc) = a + b + c - (ab + bc + ca) + abc$ | M1 A1 A1 [3] | 1.2 1.1 1.1 | For attempt at 1st side; For correctly obtaining 1st side; For legitimately obtaining 2nd side
**(ii)** | (A, ○) | 0 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 0 | 0 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 0 | 6 | 5 | 4 | 3 |
| 3 | 3 | 6 | 4 | 2 | 0 | 5 |
| 4 | 4 | 5 | 2 | 6 | 3 | 0 |
| 5 | 5 | 4 | 0 | 3 | 6 | 2 |
| 6 | 6 | 3 | 5 | 0 | 2 | 4 | M1 A1 A1 [3] | 1.1a 1.1 1.1 | General idea (≥ 1 row or column correct); If no more than six errors; All correct
**(iii)** Closure follows from table (no new elements); (Associativity follows from part (i)); 0 is the identity; 2 is self-inverse; (3, 5) and (4, 6) are inverse-pairs, so every (non-identity) element has an inverse in the set | B1 B1 B1 [3] | 2.2a 2.2a 2.1 | Told to assume this; Told to assume this; Must follow from correct table in (ii)
**(iv)** {0}, {0, 2}, {0, 4, 6}, A | B1 B1 [2] | 1.1 2.5 | 1st B for either non-trivial subgroup; 2nd B for all 4 and no extras and correct use of set/group notation4 (i) The binary operation is defined on $\mathbb { Z }$ by $a$ b $b = a + b - a b$ for all $a , b \in \mathbb { Z }$. Prove that is associative on $\mathbb { Z }$.
The operation ∘ is defined on the set $A = \{ 0,2,3,4,5,6 \}$ by $a \circ b = a + b - a b ( \bmod 7 )$ for all $a , b \in A$.\\
(ii) Complete the Cayley table for $\left( A , { } ^ { \circ } \right)$ given in the Printed Answer Booklet.\\
(iii) Prove that $( A , \circ )$ is a group. You may assume that the operation is associative.\\
(iv) List all the subgroups of $( A , \circ )$.
\hfill \mbox{\textit{OCR Further Additional Pure AS 2017 Q4 [11]}}