AQA Further AS Paper 2 Discrete 2023 June — Question 5 7 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Discrete (Further AS Paper 2 Discrete)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeComplete or analyse Cayley table
DifficultyStandard +0.3 This question tests basic group theory concepts (identity, self-inverse elements) and Cayley table completion with commutativity. Part (a) requires simple modular arithmetic checks, part (b)(i) uses commutativity to fill gaps, and part (b)(ii) asks for a counterexample to associativity—all straightforward applications of definitions with minimal problem-solving required. Slightly easier than average A-level due to the direct nature of the tasks.
Spec8.02e Finite (modular) arithmetic: integers modulo n8.02f Single linear congruences: solve ax = b (mod n)8.03a Binary operations: and their properties on given sets8.03b Cayley tables: construct for finite sets under binary operation
5
  1. The set \(S\) is defined as \(S = \{ 0,1,2,3,4,5 \}\) 5
    1. (i) State the identity element of \(S\) under the operation multiplication modulo 6 5
    2. (ii) An element \(g\) of a set is said to be self-inverse under a binary operation * if $$g * g = e$$ where \(e\) is the identity element of the set. Find all the self-inverse elements in \(S\) under the operation multiplication modulo 6
      5
    3. \(\quad\) The set \(T\) is defined as $$T = \{ a , b , c \}$$ Figure 1 shows a partially completed Cayley table for \(T\) under the commutative binary operation - \begin{table}[h]
      \captionsetup{labelformat=empty} \caption{Figure 1}
      -\(a\)\(b\)c
      \(a\)\(a\)cb
      \(b\)\(b\)\(a\)
      cc
      \end{table} 5
      1. Complete the Cayley table in Figure 1 5
    5. (ii) Prove that is not associative when acting on the elements of \(T\)
Question 5(a)(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(1\)B1 States correct identity element
Question 5(a)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(0 \times_6 0 = 0\), \(1 \times_6 1 = 1\), \(2 \times_6 2 = 4\), \(3 \times_6 3 = 3\), \(4 \times_6 4 = 4\), \(5 \times_6 5 = 1\); \(1 \times_6 1\) and \(5 \times_6 5\) produce the identity element \(1\)M1 Uses identity element to search for self-inverse elements; PI by at least one correct self-inverse element
Therefore, self-inverse elements of \(S\) are \(1\) and \(5\)A1 Correctly identifies both self-inverse elements of \(S\) and no others
Question 5(b)(i):
AnswerMarks Guidance
AnswerMarks Guidance
Cayley table: \(\bullet \mid a \mid b \mid c\); row \(a\): \(a, c, b\); row \(b\): \(c, b, a\); row \(c\): \(b, a, c\)B1 Correctly completes Cayley table
Question 5(b)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\((a \bullet b) \bullet c = c \bullet c = c\) and \(a \bullet (b \bullet c) = a \bullet a = a\)M1 Sets up a test for associativity by stating 2 appropriate expressions
As \((a \bullet b) \bullet c = c \neq a = a \bullet (b \bullet c)\)A1 Correctly evaluates their 2 appropriate expressions
Therefore \(\bullet\) is not associativeR1 Constructs a complete mathematical argument to justify non-associativity 
# Question 5(a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $1$ | B1 | States correct identity element |

---

# Question 5(a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 \times_6 0 = 0$, $1 \times_6 1 = 1$, $2 \times_6 2 = 4$, $3 \times_6 3 = 3$, $4 \times_6 4 = 4$, $5 \times_6 5 = 1$; $1 \times_6 1$ and $5 \times_6 5$ produce the identity element $1$ | M1 | Uses identity element to search for self-inverse elements; PI by at least one correct self-inverse element |
| Therefore, self-inverse elements of $S$ are $1$ and $5$ | A1 | Correctly identifies both self-inverse elements of $S$ and no others |

---

# Question 5(b)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley table: $\bullet \mid a \mid b \mid c$; row $a$: $a, c, b$; row $b$: $c, b, a$; row $c$: $b, a, c$ | B1 | Correctly completes Cayley table |

---

# Question 5(b)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(a \bullet b) \bullet c = c \bullet c = c$ and $a \bullet (b \bullet c) = a \bullet a = a$ | M1 | Sets up a test for associativity by stating 2 appropriate expressions |
| As $(a \bullet b) \bullet c = c \neq a = a \bullet (b \bullet c)$ | A1 | Correctly evaluates their 2 appropriate expressions |
| Therefore $\bullet$ is not associative | R1 | Constructs a complete mathematical argument to justify non-associativity |

---
5
\begin{enumerate}[label=(\alph*)]
\item The set $S$ is defined as $S = \{ 0,1,2,3,4,5 \}$

5 (a) (i) State the identity element of $S$ under the operation multiplication modulo 6

5 (a) (ii) An element $g$ of a set is said to be self-inverse under a binary operation * if

$$g * g = e$$

where $e$ is the identity element of the set.

Find all the self-inverse elements in $S$ under the operation multiplication modulo 6\\

5
\item $\quad$ The set $T$ is defined as

$$T = \{ a , b , c \}$$

Figure 1 shows a partially completed Cayley table for $T$ under the commutative binary operation -

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\begin{tabular}{|l|l|l|l|}
\hline
- & $a$ & $b$ & c \\
\hline
$a$ & $a$ & c & b \\
\hline
$b$ &  & $b$ & $a$ \\
\hline
c &  &  & c \\
\hline
\end{tabular}
\end{center}
\end{table}

5 (b) (i) Complete the Cayley table in Figure 1

5 (b) (ii) Prove that is not associative when acting on the elements of $T$
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2023 Q5 [7]}}
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