AS June 2023 Q1
1. The operation \(*\) is defined on the set \(G = \{0, 1, 2, 3\}\) by
\[x * y \equiv x + y - 2xy \pmod{4}\](a) Complete the Cayley table below.
(2)
| \(*\) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| 0 | ||||
| 1 | ||||
| 2 | ||||
| 3 |
(b) Show that \(G\) is a group under the operation \(*\)
(You may assume the associative law is satisfied.) (3)
(You may assume the associative law is satisfied.) (3)
(c) State the order of each element of \(G\). (2)
(d) State whether \(G\) is a cyclic group, giving a reason for your answer. (1)
| Scheme | Marks | AO | |||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 | 1.1b 1.1b | |||||||||||||||||||||||||
| (2) |
Notes
M1: Finds at least 6 correct entries
A1: All entries correct
| Scheme | Marks | AO |
|---|---|---|
| Identity element is 0 and there is closure | M1 | 2.1 |
| 0 is the identity so is self-inverse 1, 2 and 3 are self-inverse | M1 | 2.5 |
| Associative law is assumed so \(G\) forms a group | A1 | 1.1b |
| (3) |
Notes
M1: States closure and identifies 0 as the identity element.
M1: Finds inverses for each element.
A1: Scores both previous method marks and states that the associative law is satisfied therefore \(G\) is a group
| Scheme | Marks | AO |
|---|---|---|
| 0 has order 1 and 1, 2 and 3 have order 2 | M1 A1 | 1.1b 1.1b |
| (2) |
Notes
M1: States correctly the order of at least two elements
A1: States correctly the order of all four elements
| Scheme | Marks | AO |
|---|---|---|
| There is no element with order 4 therefore \(G\) is not a cyclic group Every element is its own inverse therefore no group generator therefore \(G\) is not a cyclic group | B1 | 2.4 |
| (1) | ||
| (8 marks) |
Notes
B1: See scheme