| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Complete or analyse Cayley table |
| Difficulty | Standard +0.3 This is a straightforward Cayley table question requiring identification of the identity element (reading off the table), stating an inverse (direct lookup), and comparing two group structures. The proof that Mali is incorrect simply requires noting that one structure has an identity with all elements having inverses while the other doesn't. This is below average difficulty as it involves routine table reading and basic group property checking with no complex reasoning or novel insight required. |
| 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 group8.03d Latin square property: for group tables |
| \(\odot\) | \(a\) | \(b\) | \(c\) | \(d\) |
| \(a\) | \(a\) | \(a\) | \(a\) | \(a\) |
| \(b\) | \(a\) | \(d\) | \(b\) | \(c\) |
| \(c\) | \(a\) | \(b\) | \(c\) | \(d\) |
| \(d\) | \(a\) | \(c\) | \(d\) | \(a\) |
| \(\times _ { 4 }\) | 0 | 1 | 2 | 3 |
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 |
| 2 | 0 | 2 | 0 | 2 |
| 3 | 0 | 3 | 2 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a \square\ c = c \square\ a = a\), \(b \square\ c = c \square\ b = b\), \(c \square\ c = c\), \(d \square\ c = c \square\ d = d\) — begins proof by exhaustion by identifying that all elements remain unchanged combined with \(c\) | 1.1a | M1 |
| Every element remains unchanged when combined with \(c\) under the binary operation \(\square\); therefore \(c\) is the identity element of \(S\) under the binary operation \(\square\) — completes a reasoned argument using both left and right multiplication or using commutativity | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(d\) | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| From Figure 3, the identity element is 1 | 1.1b | B1 |
| \(1 \times_4 1 = 1\), \(3 \times_4 3 = 1\); there are no pairs of distinct elements that produce the identity element — uses identity element to search for pairs of inverse elements from Figure 3, OR forms a mapping between all four elements of Figure 2 and Figure 3 | 1.1a | M1 |
| However, from part (a), the inverse of \(b\) is \(d\), a different member of the set; therefore the Cayley table in Figure 2 cannot represent multiplication modulo 4 — completes a reasoned argument to conclude that Mali's statement is incorrect | 2.1 | R1 |
# Question 8(a)(i):
$a \square\ c = c \square\ a = a$, $b \square\ c = c \square\ b = b$, $c \square\ c = c$, $d \square\ c = c \square\ d = d$ — begins proof by exhaustion by identifying that all elements remain unchanged combined with $c$ | 1.1a | M1
Every element remains unchanged when combined with $c$ under the binary operation $\square$; therefore $c$ is the identity element of $S$ under the binary operation $\square$ — completes a reasoned argument using both left and right multiplication or using commutativity | 2.1 | R1
---
# Question 8(a)(ii):
$d$ | 1.1b | B1
---
# Question 8(b):
From Figure 3, the identity element is 1 | 1.1b | B1
$1 \times_4 1 = 1$, $3 \times_4 3 = 1$; there are no pairs of distinct elements that produce the identity element — uses identity element to search for pairs of inverse elements from Figure 3, OR forms a mapping between all four elements of Figure 2 and Figure 3 | 1.1a | M1
However, from part (a), the inverse of $b$ is $d$, a different member of the set; therefore the Cayley table in Figure 2 cannot represent multiplication modulo 4 — completes a reasoned argument to conclude that Mali's statement is incorrect | 2.1 | R1
**Total: 6**
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**Paper total: 40**8 The set $S$ is defined as
$$S = \{ a , b , c , d \}$$
Figure 2 shows a Cayley table for $S$ under the commutative binary operation
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\begin{tabular}{|l|l|l|l|l|}
\hline
$\odot$ & $a$ & $b$ & $c$ & $d$ \\
\hline
$a$ & $a$ & $a$ & $a$ & $a$ \\
\hline
$b$ & $a$ & $d$ & $b$ & $c$ \\
\hline
$c$ & $a$ & $b$ & $c$ & $d$ \\
\hline
$d$ & $a$ & $c$ & $d$ & $a$ \\
\hline
\end{tabular}
\end{center}
\end{table}
8
\begin{enumerate}[label=(\alph*)]
\item (i) Prove that there exists an identity element for $S$ under the binary operation\\[0pt]
[2 marks]\\
8 (a) (ii) State the inverse of $b$ under the binary operation\\
8
\item Figure 3 shows a Cayley table for multiplication modulo 4
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\begin{tabular}{|l|l|l|l|l|}
\hline
$\times _ { 4 }$ & 0 & 1 & 2 & 3 \\
\hline
0 & 0 & 0 & 0 & 0 \\
\hline
1 & 0 & 1 & 2 & 3 \\
\hline
2 & 0 & 2 & 0 & 2 \\
\hline
3 & 0 & 3 & 2 & 1 \\
\hline
\end{tabular}
\end{center}
\end{table}
Mali says that, by substituting suitable distinct values for $a , b , c$ and $d$, the Cayley table in Figure 2 could represent multiplication modulo 4
Use your answers to part (a) to show that Mali's statement is incorrect.\\
\includegraphics[max width=\textwidth, alt={}, center]{21ed3b4e-a089-4607-b5d6-69d8aac03f31-20_2491_1736_219_139}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2020 Q8 [6]}}