| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Prove group-theoretic identities |
| Difficulty | Standard +0.3 This is a straightforward group theory question testing basic definitions and properties. Part (i) involves reading a Cayley table to identify the identity (trivial), compute an inverse (one table lookup), check closure for subgroups (routine verification). Part (ii) requires manipulating group elements using given relations and orders, which is mechanical algebra once you know to use x³=e and y⁶=e. All parts are standard textbook exercises requiring recall and direct application of definitions rather than problem-solving insight. Slightly easier than average A-level due to its procedural nature. |
| 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 group8.03f Subgroups: definition and tests for proper subgroups |
| 。 | \(a\) | \(b\) | \(c\) | \(d\) | \(e\) | \(f\) |
| \(a\) | \(d\) | c | \(b\) | \(a\) | \(f\) | \(e\) |
| \(b\) | \(e\) | \(f\) | \(a\) | \(b\) | \(c\) | \(d\) |
| \(c\) | \(f\) | \(e\) | \(d\) | \(c\) | \(b\) | \(a\) |
| \(d\) | \(a\) | \(b\) | \(c\) | \(d\) | \(e\) | \(f\) |
| \(e\) | \(b\) | \(a\) | \(f\) | \(e\) | \(d\) | \(c\) |
| \(f\) | c | \(d\) | \(e\) | \(f\) | \(a\) | \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Identity is \(d\) | B1 | Correct element, \(d\), identified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((b \circ c)^{-1} = a^{-1}\) | M1 | Correct method to find inverse of \(b \circ c\). May find \(b \circ c\) first and identify its inverse, or may apply inverse property and find \(c \circ f\). Alt: \((b \circ c)^{-1} = c^{-1} \circ b^{-1} = c \circ f\) |
| \(= a\) | A1 | Correct element, \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The set cannot be a subgroup because e.g. \(G\) has order 6 and 4 does not divide 6 so by Lagrange's Theorem, or the identity is not in the set, or closure fails e.g. \(a \circ a = d\) is not in the set | B1 | Any correct reason given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Table for \(\{b, d, f\}\): produces correct Cayley table | M1 | Investigates closure e.g. by drawing the table for \(\{b,d,f\}\) or finding individual products (at least 3) or by considering \(\langle b \rangle = \{b, b^2, b^3\} = \{b, f, d\}\) so cyclic |
| Closed, \(b\) and \(f\) are inverses and \(d\) is the identity, so the subset is a subgroup | A1 | Correctly shows closure, and refers to inverse and identity, and makes conclusion it is a subgroup |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y^3 x y^3 x^2 = y^2 y x y^3 x^2 = y^2 x y^5 y^3 x^2\) | M1 | Applies the relation \(yx = xy^5\) at least once to change positions of an \(x\) and a \(y\). May work from either end |
| \(= y y x y^8 x^2 = y x y^5 y^8 x^2 = x y^5 y^{13} x^2\) | M1 | Completes the process of gathering all \(y\)'s or \(x\)'s together by using the relation repeatedly; may reduce elements according to order throughout |
| \(= x y^{18} x^2 = x e x^2 = x^3 = e\) * | A1\* | Uses the orders of the elements to reduce gathered \(y\) terms and \(x\) terms down to identity to complete the proof |
# Question 1:
## Part (i)(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Identity is $d$ | **B1** | Correct element, $d$, identified |
**(1 mark)**
---
## Part (i)(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(b \circ c)^{-1} = a^{-1}$ | **M1** | Correct method to find inverse of $b \circ c$. May find $b \circ c$ first and identify its inverse, or may apply inverse property and find $c \circ f$. Alt: $(b \circ c)^{-1} = c^{-1} \circ b^{-1} = c \circ f$ |
| $= a$ | **A1** | Correct element, $a$ |
**(2 marks)**
---
## Part (i)(c)
| Answer | Mark | Guidance |
|--------|------|----------|
| The set cannot be a subgroup because e.g. $G$ has order 6 and 4 does not divide 6 so by Lagrange's Theorem, or the identity is not in the set, or closure fails e.g. $a \circ a = d$ is not in the set | **B1** | Any correct reason given |
**(1 mark)**
---
## Part (i)(d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Table for $\{b, d, f\}$: produces correct Cayley table | **M1** | Investigates closure e.g. by drawing the table for $\{b,d,f\}$ or finding individual products (at least 3) or by considering $\langle b \rangle = \{b, b^2, b^3\} = \{b, f, d\}$ so cyclic |
| Closed, $b$ and $f$ are inverses and $d$ is the identity, so the subset is a subgroup | **A1** | Correctly shows closure, and refers to inverse and identity, and makes conclusion it is a subgroup |
**(2 marks)**
---
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y^3 x y^3 x^2 = y^2 y x y^3 x^2 = y^2 x y^5 y^3 x^2$ | **M1** | Applies the relation $yx = xy^5$ at least once to change positions of an $x$ and a $y$. May work from either end |
| $= y y x y^8 x^2 = y x y^5 y^8 x^2 = x y^5 y^{13} x^2$ | **M1** | Completes the process of gathering all $y$'s or $x$'s together by using the relation repeatedly; may reduce elements according to order throughout |
| $= x y^{18} x^2 = x e x^2 = x^3 = e$ * | **A1\*** | Uses the orders of the elements to reduce gathered $y$ terms and $x$ terms down to identity to complete the proof |
**(3 marks)**
---
**Total: 9 marks**\begin{enumerate}
\item (i) The table below is a Cayley table for the group $G$ with operation ∘
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
。 & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \\
\hline
$a$ & $d$ & c & $b$ & $a$ & $f$ & $e$ \\
\hline
$b$ & $e$ & $f$ & $a$ & $b$ & $c$ & $d$ \\
\hline
$c$ & $f$ & $e$ & $d$ & $c$ & $b$ & $a$ \\
\hline
$d$ & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \\
\hline
$e$ & $b$ & $a$ & $f$ & $e$ & $d$ & $c$ \\
\hline
$f$ & c & $d$ & $e$ & $f$ & $a$ & $b$ \\
\hline
\end{tabular}
\end{center}
(a) State which element is the identity of the group.\\
(b) Determine the inverse of the element ( $b \circ c$ )\\
(c) Give a reason why the set $\{ a , b , e , f \}$ cannot be a subgroup of $G$. You must justify your answer.\\
(d) Show that the set $\{ b , d , f \}$ is a subgroup of $G$.\\
(ii) Given that $H$ is a group with an element $x$ of order 3 and an element $y$ of order 6 satisfying
$$y x = x y ^ { 5 }$$
show that $y ^ { 3 } x y ^ { 3 } x ^ { 2 }$ is the identity element.\\
\includegraphics[max width=\textwidth, alt={}, center]{7d269bf1-f481-46bd-b9d3-fea211b186cf-02_2270_54_309_1980}
\hfill \mbox{\textit{Edexcel FP2 AS 2024 Q1 [9]}}