A2 June 2025 Q1
1. The set \(S = \{1, 3, 5, 9, 11, 13\}\) forms the group \(G\), under the operation multiplication modulo 14
| \(\times_{14}\) | 1 | 3 | 5 | 9 | 11 | 13 |
|---|---|---|---|---|---|---|
| 1 | 1 | 3 | 5 | 9 | 11 | 13 |
| 3 | 3 | 9 | 1 | 13 | 5 | 11 |
| 5 | 5 | 1 | 11 | |||
| 9 | 9 | 13 | 11 | |||
| 11 | 11 | 5 | 9 | |||
| 13 | 13 | 11 | 1 |
The group \(H\) is defined by the Cayley table below.
| \(*\) | \(p\) | \(q\) | \(r\) | \(s\) | \(t\) | \(u\) |
|---|---|---|---|---|---|---|
| \(p\) | \(p\) | \(q\) | \(r\) | \(s\) | \(t\) | \(u\) |
| \(q\) | \(q\) | \(t\) | \(u\) | \(r\) | \(s\) | \(p\) |
| \(r\) | \(r\) | \(u\) | \(t\) | \(q\) | \(p\) | \(s\) |
| \(s\) | \(s\) | \(r\) | \(q\) | \(p\) | \(u\) | \(t\) |
| \(t\) | \(t\) | \(s\) | \(p\) | \(u\) | \(r\) | \(q\) |
| \(u\) | \(u\) | \(p\) | \(s\) | \(t\) | \(q\) | \(r\) |
| Scheme | Marks | AO | |||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 A1 | 1.1b 1.1b 1.1b | |||||||||||||||||||||||||||||||||||||||||||||||||
| (3) |
Notes
M1: At least 3 correct entries.
A1: At least 6 correct entries.
A1: All 12 entries correct.
NB for the final A all entries must be given modulo 14. Do not accept e.g. 17 for 3.
| Scheme | Marks | AO |
|---|---|---|
| \(\{1, 13\}\) | B1 | 2.2a |
| (1) |
Notes
B1: Deduces the correct subgroup. This is the only subgroup of order 2. Accept with set brackets or round brackets as long as both elements are listed. Accept \(\langle 13 \rangle\) as long as the angled brackets are clear.
| Scheme | Marks | AO |
|---|---|---|
| States the groups are cyclic or identifies a generator for each. E.g. \(3^2 = 9, 3^3 = 13, 3^4 = 11, 3^5 = 5, 3^6 = 1\) so 3 generates G or \(q^2 = t, q^3 = s, q^4 = r, q^5 = u, q^6 = p\) so \(q\) generates \(H\) | M1 | 3.1a |
| E.g. \((3^2 = 9, 3^3 = 13, 3^4 = 11, 3^5 = 5, 3^6 = 1)\) 3 (or 5) generates G and \((q^2 = t, q^3 = s, q^4 = r, q^5 = u, q^6 = p)\) \(q\) (or u) generates \(H\) | A1 | 1.1b |
| Both groups are cyclic of order 6 and so are isomorphic. | A1 | 2.1 |
| (3) | ||
| (7 marks) |
Notes
M1: State the groups are cyclic, or identifies a generator for each group.
A1: Correctly generator identified for each group.
A1: Correct work for showing a generator for each group and conclude both cyclic and hence isomorphic.
Alt I
| Scheme | Marks | AO | |||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 | 3.1a 1.1b | |||||||||||||||||||||
| Also \(3^2 = 9\) and \(\theta(3)^2 = q^2 = t = \theta(9) = \theta(3^2)\) so structure is preserved and therefore groups are isomorphic | A1 | 2.1 | |||||||||||||||||||||
| (3) |
M1: Attempts to identify an isomorphism between the elements – may be implied by
- identifying at least 4 correct pairings
- by attempting to rearrange group tables to have the same structure
A1: For a correct pairing of all items for an isomorphism. Note that \(3^2 = 9\) while \(q^2 = t\) and this can be used to see if they have a fully correct pairing (check if \(3 \leftrightarrow q\) then you need \(9 \leftrightarrow t\), or if \(3 \leftrightarrow u\) then you need \(9 \leftrightarrow r\))
A1: All correct pairings with demonstration of structure being preserved for at least one of the possible choices made, or via matching of group tables to show same structure, and concludes isomorphic.
NB there are two choices for isomorphism, \(1 \leftrightarrow p\) and \(13 \leftrightarrow s\) are forced, 3 and 5 must map to \(q\) and \(u\) either way round, but then 9 and 11 need to be correctly selected for the order of 3 and 5 – and this must be shown for fully marks. They may likely choose the pairings by considering orders (as in Alt II) but still need a reason for the final choice.
Alt II
| Scheme | Marks | AO | ||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 | 3.1a 1.1b | ||||||||||||||||||||||||||||
| Since orders match and group is order 6 (so one of only two types) therefore the groups are isomorphic isomorphic | A1 | 2.1 | ||||||||||||||||||||||||||||
| (3) |
M1: Attempts to find the order for each element of a both groups – at least four correct for at least one of the groups.
A1: All orders correct for all elements of both groups.
A1: All orders correct, reason referring to small group order and therefore groups are isomorphic. Note if they go on to give an incorrect pairing it is wrong reasoning so A0.
Note: In general matching orders of all elements is a necessary but not sufficient requirement for an isomorphism (e.g. for groups of order 16 there are distinct isomorphism classes which have the same number of elements of the same orders). However, for small groups (up to order 15) with few isomorphism classes this criterion is sufficient but for a fully correct answer some reference to the group order must be given.