A2 June 2019 Q6
6.
The set \(H = \{1, 5, 7, 11, 13, 17\}\) forms a group under the operation of multiplication modulo 18
| Scheme | Marks | AO |
|---|---|---|
| \((a\mathbin{*}b)\mathbin{*}c = (a + b + ab)\mathbin{*}c = a + b + ab + c + (a + b + ab)c\) | M1 | 2.1 |
| \(a\mathbin{*}(b\mathbin{*}c) = a\mathbin{*}(b + c + bc) = a + b + c + bc + a(b + c + bc)\) | M1 | 2.1 |
| \(\underline{a + b + ab + c + (a + b + ab)c} = a + b + c + bc + ab + ac + abc\) \(\phantom{a + b + ab + c + (a + b + ab)c} = \underline{a + b + c + bc + a(b + c + bc)}\) | A1 | 2.2a |
| so \((a\mathbin{*}b)\mathbin{*}c = a\mathbin{*}(b\mathbin{*}c)\) which means \(*\) is associative | A1 | 2.4 |
| (4) |
Notes
M1: Begins proof by correctly expanding \((a\mathbin{*}b)\mathbin{*}c\) or \(a\mathbin{*}(b\mathbin{*}c)\) to an expression in \(a\), \(b\) and \(c\). Note they may expand as \((a\mathbin{*}b)\mathbin{*}c = (a\mathbin{*}b) + c + (a\mathbin{*}b)c = a + b + ab + c + (a + b + ab)c\) which is equally fine.
M1: Makes progress towards the required result by attempting to expand both \((a\mathbin{*}b)\mathbin{*}c\) and \(a\mathbin{*}(b\mathbin{*}c)\), but be generous with the attempts for this method. May achieve this by working from left to right, so look for arriving at the other expression through a chain of equalities.
A1: For both underlined expressions (but accept eg. \(c(a + b + ab)\) for \((a + b + ab)c\)) and a correct expansion seen for each independently or part of a chain as shown. The expansion may have terms in different orders.
A1: Explains that \((a\mathbin{*}b)\mathbin{*}c = a\mathbin{*}(b\mathbin{*}c)\) means that \(*\) is associative. Depends on both M marks and a correct expression having been found.
| Scheme | Marks | AO |
|---|---|---|
| \(3^2 = 2\quad 3^3 = 6\quad 3^4 = 4\quad 3^5 = 5\quad 3^6 = 1\) or \(5^2 = 4\quad 5^3 = 6\quad 5^4 = 2\quad 5^5 = 3\quad 5^6 = 1\) Or special case for M1A0 if powers not shown: 3 has order 6 so generates the group | M1 | 2.1 |
| 3 (or 5) has order 6 and so generates the group so \(G\) is cyclic | A1 | 2.4 |
| (2) |
Notes
M1: Demonstrates understanding of the term cyclic by either attempting all the powers of 3 or 5. Accept for this a statement \(\langle 3\rangle = \{3, 2, 6, 4, 5, 1\}\) which shows the elements list in order of powers.
A1: Must have evaluated all powers of 3 or 5 correctly and explains why the group is cyclic. Accept as 3 generates the group, or as 3 has the same order of \(G\) as reason. Must refer to cyclic in conclusion.
Special case: Allow M1A0 for a correct explanation of why \(G\) is cyclic if the order of 3 (or 5) is stated as 6 without justification – but must include reference to either being a generator or having the same order as \(G\).
| Scheme | Marks | AO |
|---|---|---|
| \(\{1\}\), \(H\) | B1 | 1.1b |
| \(\{1, 17\}\) or \(\{1, 7, 13\}\) | M1 | 1.1b |
| \(\{1, 17\}\) and \(\{1, 7, 13\}\) (and no others) | A1 | 1.1b |
| (3) |
Notes
(You may ignore references to the operation for this part)
B1: Identifies \(\{1\}\) and \(H\) as subgroups
M1: Identifies \(\{1, 17\}\) or \(\{1, 7, 13\}\) as a subgroup
A1: Identifies \(\{1, 17\}\) and \(\{1, 7, 13\}\) as subgroups and no others
| Scheme | Marks | AO | ||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 A1 | 3.1a 1.1b 1.1b | ||||||||||||||||||||||||||||
| (3) | ||||||||||||||||||||||||||||||
| (12 marks) |
Notes
M1: Attempts to identify an isomorphism between the groups – may be implied by
- identifying at least 2 correct non-identity pairings or
- by attempting to rearrange group tables to have the same structure, or
- by attempting to map powers of a generator to powers of a generator e.g \((\text{their } 3)^k \to (\text{their } 5)^k\) or
- by matching of non-trivial proper subgroups to each other.
A1: Identifies 4 correct pairings, or sets up a mapping with one correct generator
A1: All pairings correct, or sets up a mapping with generators of each group correct, eg. \(3^k \to 5^k\)