| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Complete or analyse Cayley table |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question on group theory basics. Part (a) requires simple modular arithmetic (multiplication mod 4). Part (b)(i) uses commutativity to fill in a small Cayley table by symmetry. Part (b)(ii) requires checking associativity, which is systematic but tedious for a 4-element set. While this is Further Maths content, it's mostly mechanical application of definitions with minimal conceptual insight required, making it slightly easier than average overall. |
| Spec | 8.02e Finite (modular) arithmetic: integers modulo n8.03a 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 group |
| • | \(a\) | \(b\) | \(c\) | \(d\) |
| \(a\) | \(b\) | \(a\) | \(a\) | \(c\) |
| \(b\) | \(c\) | \(c\) | ||
| \(c\) | \(d\) | \(d\) | ||
| \(d\) | \(d\) | \(d\) |
| Answer | Marks |
|---|---|
| Partially completes Cayley table (\(\times_4\)) with 3 correct rows or columns | M1 |
| Fully correct table: rows 0,1,2,3 giving \((0,0,0,0)\), \((0,1,2,3)\), \((0,2,0,2)\), \((0,3,2,1)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Completes operation table for \(S = \{*, a, b, c, d\}\) correctly | B1 | Full table required |
| Answer | Marks | Guidance |
|---|---|---|
| Sets up test for associativity with at least two different elements of \(S\): \(a \bullet (b \bullet c) = a \bullet d = c\) and \((a \bullet b) \bullet c = a \bullet c = a\) | M1 | Must use at least two different elements |
| \(a \bullet (b \bullet c) \neq (a \bullet b) \bullet c\), therefore \(\bullet\) is not associative | R1 | Complete mathematical argument required |
# Question 5:
## Part 5(a):
| Partially completes Cayley table ($\times_4$) with 3 correct rows or columns | M1 | |
| Fully correct table: rows 0,1,2,3 giving $(0,0,0,0)$, $(0,1,2,3)$, $(0,2,0,2)$, $(0,3,2,1)$ | A1 | |
## Part 5(b)(i):
| Completes operation table for $S = \{*, a, b, c, d\}$ correctly | B1 | Full table required |
## Part 5(b)(ii):
| Sets up test for associativity with at least two different elements of $S$: $a \bullet (b \bullet c) = a \bullet d = c$ and $(a \bullet b) \bullet c = a \bullet c = a$ | M1 | Must use at least two different elements |
| $a \bullet (b \bullet c) \neq (a \bullet b) \bullet c$, therefore $\bullet$ is not associative | R1 | Complete mathematical argument required |
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5
\begin{enumerate}[label=(\alph*)]
\item Complete the Cayley table in Figure 1 for multiplication modulo 4
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{dcf97b92-d067-41d4-89a6-ea5bab9ea4ff-08_761_1017_434_493}
\end{center}
\end{figure}
5
\item The set $S$ is defined as
$$S = \{ a , b , c , d \}$$
Figure 2 shows an incomplete 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
• & $a$ & $b$ & $c$ & $d$ \\
\hline
$a$ & $b$ & $a$ & $a$ & $c$ \\
\hline
$b$ & & $c$ & & $c$ \\
\hline
$c$ & & $d$ & $d$ & \\
\hline
$d$ & & & $d$ & $d$ \\
\hline
\end{tabular}
\end{center}
\end{table}
5 (b) (i) Complete the Cayley table in Figure 2.
5 (b) (ii) Determine whether the binary operation • is associative when acting on the elements of $S$.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2019 Q5 [5]}}