AQA Further Paper 3 Discrete Specimen — Question 5 10 marks

Exam BoardAQA
ModuleFurther Paper 3 Discrete (Further Paper 3 Discrete)
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeVerify group axioms
DifficultyStandard +0.8 This is a Further Maths group theory question requiring verification of four group axioms (closure, associativity, identity, inverses) for a non-standard operation, finding subgroups, and testing isomorphism. While systematic, it demands understanding of abstract algebra concepts beyond standard A-level and careful verification of multiple properties, placing it moderately above average difficulty.
Spec8.03a 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

5 The binary operation * is defined as $$a * b = a + b + 4 ( \bmod 6 )$$ where \(a , b \in \mathbb { Z }\). 5
  1. Show that the set \(\{ 0,1,2,3,4,5 \}\) forms a group \(G\) under *.
    5
  2. Find the proper subgroups of the group \(G\) in part (a).
    5
  3. Determine whether or not the group \(G\) in part (a) is isomorphic to the group \(K = \left( \langle 3 \rangle , \times _ { 14 } \right)\) [0pt] [3 marks]

Question 5(a):
AnswerMarks Guidance
AnswerMark Guidance
As all answers to \(a * b\) are reduced modulo 6, they are in the given set and thus the set is closed under \(*\)R1 Must show that under modulo 6, \(a * b\) can only result in a member of the given set
Identity element \(= 2\)B1 Clearly identifies the identity element
0 and 4 are inverses of each other; 1 and 3 are inverses of each other; 2 and 5 are self-inverse elementsB1 Finds and states the inverse of each element (PI), FT from 'their' identity
\(a*(b*c) = a+(b+c+4)+4\) is shown to equal \((a*b)*c = (a+b+4)+c+4\)R1 Shows associativity between elements of the set under the operation \(*\)
As \(G\) satisfies each of the four group axioms under the binary operation \(*\), \(G\) is a groupR1 States correct conclusion; only award if completely correct solution, clear and no slips
Question 5(b):
AnswerMarks Guidance
AnswerMark Guidance
\(\{2\}, \{0,2,4\}, \{2,5\}\) — identifies two correct subgroupsB1 Condone inclusion of \(\{0,1,2,3,4,5\}\)
Identifies all three proper subgroups and no others includedB1
Question 5(c):
AnswerMarks Guidance
AnswerMark Guidance
\(G = (\langle 1 \rangle, *)\) OR \(K = (\{3,9,13,11,5,1\}, \times_{14})\)B1 Identifies the generator of \(G\) OR generates every element of the group \(K\) (PI)
\(1 \mapsto 3,\ 0 \mapsto 9,\ 5 \mapsto 13,\ 4 \mapsto 11,\ 3 \mapsto 5,\ 2 \mapsto 1\)B1 Finds correctly a one-to-one mapping between each element of \(G\) and \(K\) (condone use of equal sign)
As there is a one-to-one mapping between the elements of \(G\) and the elements of \(K\), \(G \cong K\)E1 Deduces that \(G\) is isomorphic to \(K\) with concluding statement using correct mathematical language and notation throughout
## Question 5(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| As all answers to $a * b$ are reduced modulo 6, they are in the given set and thus the set is closed under $*$ | R1 | Must show that under modulo 6, $a * b$ can only result in a member of the given set |
| Identity element $= 2$ | B1 | Clearly identifies the identity element |
| 0 and 4 are inverses of each other; 1 and 3 are inverses of each other; 2 and 5 are self-inverse elements | B1 | Finds and states the inverse of each element (PI), FT from 'their' identity |
| $a*(b*c) = a+(b+c+4)+4$ is shown to equal $(a*b)*c = (a+b+4)+c+4$ | R1 | Shows associativity between elements of the set under the operation $*$ |
| As $G$ satisfies each of the four group axioms under the binary operation $*$, $G$ is a group | R1 | States correct conclusion; only award if completely correct solution, clear and no slips |

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## Question 5(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\{2\}, \{0,2,4\}, \{2,5\}$ — identifies two correct subgroups | B1 | Condone inclusion of $\{0,1,2,3,4,5\}$ |
| Identifies all three proper subgroups and no others included | B1 | |

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## Question 5(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $G = (\langle 1 \rangle, *)$ OR $K = (\{3,9,13,11,5,1\}, \times_{14})$ | B1 | Identifies the generator of $G$ OR generates every element of the group $K$ (PI) |
| $1 \mapsto 3,\ 0 \mapsto 9,\ 5 \mapsto 13,\ 4 \mapsto 11,\ 3 \mapsto 5,\ 2 \mapsto 1$ | B1 | Finds correctly a one-to-one mapping between each element of $G$ and $K$ (condone use of equal sign) |
| As there is a one-to-one mapping between the elements of $G$ and the elements of $K$, $G \cong K$ | E1 | Deduces that $G$ is isomorphic to $K$ with concluding statement using correct mathematical language and notation throughout |

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5 The binary operation * is defined as

$$a * b = a + b + 4 ( \bmod 6 )$$

where $a , b \in \mathbb { Z }$.

5
\begin{enumerate}[label=(\alph*)]
\item Show that the set $\{ 0,1,2,3,4,5 \}$ forms a group $G$ under *.\\

5
\item Find the proper subgroups of the group $G$ in part (a).\\

5
\item Determine whether or not the group $G$ in part (a) is isomorphic to the group $K = \left( \langle 3 \rangle , \times _ { 14 } \right)$\\[0pt]
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Discrete  Q5 [10]}}