| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Standard +0.3 This is a straightforward group theory question testing basic definitions and Lagrange's theorem. Parts (a)-(c) require only direct application of definitions (identity, inverse, associativity verification through computation), while part (d) is immediate recall of Lagrange's theorem. All parts are routine for Further Maths students with no problem-solving or insight required, making it slightly easier than average. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03k Lagrange's theorem: order of subgroup divides order of group |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\{e =\} \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix}\) | B1 | See scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 1 & 2 \end{pmatrix}\) | B1 | See scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Shows two calculations: \([a \circ b] \circ c\) and \(a \circ [b \circ c]\) with intermediate working using permutations, both yielding \(\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}\) | M1 | Must show both \([a \circ b] \circ c\) and \(a \circ [b \circ c]\) with an intermediate line of working. Condone wrong order for M1. |
| Correct calculations leading to \([a \circ b] \circ c = a \circ [b \circ c]\), or states associative | A1 | Note: incorrect order scores M1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The order of the group is \(24\) or \(4!\) | B1 | See scheme |
| \(4\) is a factor of \(24\), or \(4/24\), therefore it is possible for a subgroup to have order \(4\) | B1ft | Follow through on their order of group |
## Question 1:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\{e =\} \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix}$ | B1 | See scheme |
---
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 1 & 2 \end{pmatrix}$ | B1 | See scheme |
---
### Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Shows two calculations: $[a \circ b] \circ c$ and $a \circ [b \circ c]$ with intermediate working using permutations, both yielding $\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}$ | M1 | Must show both $[a \circ b] \circ c$ and $a \circ [b \circ c]$ with an intermediate line of working. Condone wrong order for M1. |
| Correct calculations leading to $[a \circ b] \circ c = a \circ [b \circ c]$, or states associative | A1 | Note: incorrect order scores M1 A0 |
---
### Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| The order of the group is $24$ or $4!$ | B1 | See scheme |
| $4$ is a factor of $24$, or $4/24$, therefore it is **possible** for a subgroup to have order $4$ | B1ft | Follow through on their order of group |\begin{enumerate}
\item 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.
\end{enumerate}
For the group $\mathrm { S } _ { 4 }$\\
(a) write down the identity element,\\
(b) write down the inverse of the element $a$, where
$$a = \left( \begin{array} { l l l l }
1 & 2 & 3 & 4 \\
3 & 4 & 2 & 1
\end{array} \right)$$
(c) demonstrate that the operation of composition is associative using the following elements
$$a = \left( \begin{array} { l l l l }
1 & 2 & 3 & 4 \\
3 & 4 & 2 & 1
\end{array} \right) \quad b = \left( \begin{array} { l l l l }
1 & 2 & 3 & 4 \\
2 & 4 & 3 & 1
\end{array} \right) \quad \text { and } c = \left( \begin{array} { l l l l }
1 & 2 & 3 & 4 \\
4 & 1 & 2 & 3
\end{array} \right)$$
(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.
\hfill \mbox{\textit{Edexcel FP2 2022 Q1 [6]}}