| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Subgroups and cosets |
| Difficulty | Challenging +1.2 Part (i) is a direct application of Lagrange's theorem requiring students to recognize that 11 doesn't divide 5291848 (without calculation) and construct a simple proof by contradiction. Part (ii) involves completing a partially-filled Cayley table through modular arithmetic calculations and checking group axioms. While this is Further Maths content, both parts are relatively routine applications of standard group theory results with no novel insight required—the Cayley table is computational and the Lagrange application is textbook. |
| Spec | 8.03a Binary operations: and their properties on given sets8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group8.03k Lagrange's theorem: order of subgroup divides order of group |
| \(\times _ { 30 }\) | 2 | 4 | 8 | 14 | 16 | 22 | 26 | 28 |
| 2 | 4 | 8 | 16 | 28 | 2 | 14 | 22 | 26 |
| 4 | 8 | 2 | 28 | 14 | ||||
| 8 | 16 | 2 | 8 | 14 | ||||
| 14 | 28 | 22 | 16 | 8 | 4 | |||
| 16 | 2 | 4 | 14 | 16 | ||||
| 22 | 14 | 26 | 4 | 2 | 16 | |||
| 26 | 22 | 14 | 4 | 8 | ||||
| 28 | 26 | 14 | 28 | 8 |
| \(\times _ { 30 }\) | 2 | 4 | 8 | 14 | 16 | 22 | 26 | 28 |
| 2 | 4 | 8 | 16 | 28 | 2 | 14 | 22 | 26 |
| 4 | 8 | 2 | 28 | 14 | ||||
| 8 | 16 | 2 | 8 | 14 | ||||
| 14 | 28 | 22 | 16 | 8 | 4 | |||
| 16 | 2 | 4 | 14 | 16 | ||||
| 22 | 14 | 26 | 4 | 2 | 16 | |||
| 26 | 22 | 14 | 4 | 8 | ||||
| 28 | 26 | 14 | 28 | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Suppose \(G\) has a subgroup of order 11, then (by Lagrange's Theorem) 11 must divide \(5291848\) | M1 | 2.1 |
| \(5-2+9-1+8-4+8=23\) | M1 | 1.1b |
| 23 is not divisible by 11, hence 11 does not divide \( | G | \), which contradicts Lagrange's Theorem. Hence there is no subgroup of order 11. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Cayley table for \((X, \times_{30})\) where \(X=\{2,4,8,14,16,22,26,28\}\) — completes at least one row or column correctly | M1 | 1.1b |
| At least 5 rows or columns completed correctly | A1 | 1.1b |
| Completely correct table | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| As the row and column for 16 repeat the borders, 16 is an identity element for \((X, \times_{30})\) | B1 | 2.2a |
| Each element has an inverse: \(2\leftrightarrow8\), \(4\leftrightarrow4\), \(8\leftrightarrow2\), \(14\leftrightarrow14\), \(16\leftrightarrow16\), \(22\leftrightarrow28\), \(26\leftrightarrow26\), \(28\leftrightarrow22\) | B1 | 1.1b |
| Since \(\times_{30}\) is associative and there are no new elements in the table, \((X,\times_{30})\) is closed, hence \((X,\times_{30})\) is a group | B1 | 2.4 |
## Question 3(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Suppose $G$ has a subgroup of order 11, then (by Lagrange's Theorem) 11 must divide $5291848$ | M1 | 2.1 |
| $5-2+9-1+8-4+8=23$ | M1 | 1.1b |
| 23 is not divisible by 11, hence 11 does not divide $|G|$, which contradicts Lagrange's Theorem. Hence there is no subgroup of order 11. | A1 | 2.4 |
---
## Question 3(ii)(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Cayley table for $(X, \times_{30})$ where $X=\{2,4,8,14,16,22,26,28\}$ — completes at least one row or column correctly | M1 | 1.1b |
| At least 5 rows or columns completed correctly | A1 | 1.1b |
| Completely correct table | A1 | 1.1b |
---
## Question 3(ii)(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| As the row and column for 16 repeat the borders, 16 is an identity element for $(X, \times_{30})$ | B1 | 2.2a |
| Each element has an inverse: $2\leftrightarrow8$, $4\leftrightarrow4$, $8\leftrightarrow2$, $14\leftrightarrow14$, $16\leftrightarrow16$, $22\leftrightarrow28$, $26\leftrightarrow26$, $28\leftrightarrow22$ | B1 | 1.1b |
| Since $\times_{30}$ is associative and there are no new elements in the table, $(X,\times_{30})$ is closed, hence $(X,\times_{30})$ is a group | B1 | 2.4 |
---\begin{enumerate}
\item (i) Let $G$ be a group of order 5291848
\end{enumerate}
Without performing any division, use proof by contradiction to show that $G$ cannot have a subgroup of order 11\\
(ii) (a) Complete the following Cayley table for the set $X = \{ 2,4,8,14,16,22,26,28 \}$ with the operation of multiplication modulo 30
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
$\times _ { 30 }$ & 2 & 4 & 8 & 14 & 16 & 22 & 26 & 28 \\
\hline
2 & 4 & 8 & 16 & 28 & 2 & 14 & 22 & 26 \\
\hline
4 & 8 & & 2 & & & 28 & 14 & \\
\hline
8 & 16 & 2 & & & 8 & & & 14 \\
\hline
14 & 28 & & 22 & 16 & & 8 & 4 & \\
\hline
16 & 2 & 4 & & 14 & 16 & & & \\
\hline
22 & 14 & & 26 & & & 4 & 2 & 16 \\
\hline
26 & 22 & 14 & & 4 & & & & 8 \\
\hline
28 & 26 & & 14 & & 28 & & 8 & \\
\hline
\end{tabular}
\end{center}
A copy of this table is given on page 11 if you need to rewrite your Cayley table.\\
(b) Hence determine whether the set $X$ with the operation of multiplication modulo 30 forms a group.\\[0pt]
[You may assume multiplication modulo $n$ is an associative operation.]
Only use this grid if you need to rewrite your Cayley table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
$\times _ { 30 }$ & 2 & 4 & 8 & 14 & 16 & 22 & 26 & 28 \\
\hline
2 & 4 & 8 & 16 & 28 & 2 & 14 & 22 & 26 \\
\hline
4 & 8 & & 2 & & & 28 & 14 & \\
\hline
8 & 16 & 2 & & & 8 & & & 14 \\
\hline
14 & 28 & & 22 & 16 & & 8 & 4 & \\
\hline
16 & 2 & 4 & & 14 & 16 & & & \\
\hline
22 & 14 & & 26 & & & 4 & 2 & 16 \\
\hline
26 & 22 & 14 & & 4 & & & & 8 \\
\hline
28 & 26 & & 14 & & 28 & & 8 & \\
\hline
\end{tabular}
\end{center}
(Total for Question 3 is 9 marks)\\
\hfill \mbox{\textit{Edexcel FP2 AS 2022 Q3 [9]}}