A2 June 2022 Q1
1. The group \(\mathrm{S}_4\) is the set of all possible permutations that can be performed on the four numbers 1, 2, 3 and 4, under the operation of composition.
For the group \(\mathrm{S}_4\)
(a) write down the identity element, (1)
(b) write down the inverse of the element \(a\), where\[a = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix}\] (1)
(c) demonstrate that the operation of composition is associative using the following elements\[a = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \quad b = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix} \quad \text{and } c = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\] (2)
(d) Explain why it is possible for the group \(\mathrm{S}_4\) to have a subgroup of order 4
You do not need to find such a subgroup. (2)
You do not need to find such a subgroup. (2)
| Scheme | Marks | AO |
|---|---|---|
| \(\{e =\}\ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix}\) | B1 | 1.1b |
| (1) |
Notes
B1: See scheme
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 1 & 2 \end{pmatrix}\) | B1 | 1.1b |
| (1) |
Notes
B1: See scheme
| Scheme | Marks | AO |
|---|---|---|
| Demonstrates that, for example: \([a \circ b] \circ c = \left[\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\right] \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\) \(= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}\) \(a \circ [b \circ c] = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \left[\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\right]\) \(= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}\) | M1 | 2.1 |
| So \([a \circ b] \circ c = a \circ [b \circ c]\) or associative | A1 | 2.4 |
| (2) |
Notes
M1: Shows two calculations in an attempt to show associative, e,g, \([a \circ b] \circ c\) and \(a \circ [b \circ c]\). There must be an intermediate line of working with evidence of using the permutations. Condone the wrong order for this mark.
A1: Correct calculations leading to \([a \circ b] \circ c = a \circ [b \circ c]\) or states associative
Note Incorrect order scores M1 A0
\[[a \circ b] \circ c = \left[\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\right] \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\]\[= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\]\[= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 2 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| The order of the group is 24 or 4! | B1 | 1.1b |
| 4 is a factor of 24 or 4/24 therefore it is possible for a subgroup to have order 4. | B1ft | 2.4 |
| (2) | ||
| (6 marks) |
Notes
B1: Order is 24 or 4!
B1ft: Follow through on their order of the group, draws the correct conclusion