Edexcel FP2 AS 2023 June — Question 1 8 marks

Exam BoardEdexcel
ModuleFP2 AS (Further Pure 2 AS)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeVerify group axioms
DifficultyStandard +0.3 This is a standard group verification question requiring systematic completion of a Cayley table and checking closure, identity, and inverses. While it involves modular arithmetic and multiple parts, the steps are routine and algorithmic with no novel insight required. The associativity assumption removes the hardest verification. Slightly easier than average due to the small finite set and straightforward calculations.
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 group8.03e Order of elements: and order of groups
  1. The operation * is defined on the set \(G = \{ 0,1,2,3 \}\) by
$$x ^ { * } y \equiv x + y - 2 x y ( \bmod 4 )$$
  1. Complete the Cayley table below.
    *0123
    0
    1
    2
    3
  2. Show that \(G\) is a group under the operation *
    (You may assume the associative law is satisfied.)
  3. State the order of each element of \(G\).
  4. State whether \(G\) is a cyclic group, giving a reason for your answer.
Question 1:
Part (a)
AnswerMarks Guidance
AnswerMarks Guidance
Cayley table completed correctly: \(*\) operation on \(\{0,1,2,3\}\) where \(a*b = \a-b\ \) (or equivalent)
All entries correct (symmetric table with diagonal all 0)A1 All entries correct
(2)
Part (b)
AnswerMarks Guidance
AnswerMarks Guidance
Identity element is \(0\) and there is closureM1 States closure and identifies \(0\) as the identity element
\(0\) is self-inverse; \(1, 2\) and \(3\) are self-inverseM1 Finds inverses for each element
Associative law is assumed so \(G\) forms a groupA1 Scores both M marks and states associative law is satisfied therefore \(G\) is a group
(3)
Part (c)
AnswerMarks Guidance
AnswerMarks Guidance
\(0\) has order \(1\) and \(1, 2\) and \(3\) have order \(2\)M1 States correctly the order of at least two elements
A1States correctly the order of all four elements
(2)
Part (d)
AnswerMarks Guidance
AnswerMarks Guidance
There is no element with order \(4\) therefore \(G\) is not a cyclic group; every element is its own inverse therefore no group generator therefore \(G\) is not a cyclic groupB1 See scheme
(1)
Total: 8 marks
# Question 1:

## Part (a)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley table completed correctly: $*$ operation on $\{0,1,2,3\}$ where $a*b = \|a-b\|$ (or equivalent) | M1 | Finds at least 6 correct entries |
| All entries correct (symmetric table with diagonal all 0) | A1 | All entries correct |
| | **(2)** | |

## Part (b)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Identity element is $0$ and there is closure | M1 | States closure and identifies $0$ as the identity element |
| $0$ is self-inverse; $1, 2$ and $3$ are self-inverse | M1 | Finds inverses for each element |
| Associative law is assumed so $G$ forms a group | A1 | Scores both M marks and states associative law is satisfied therefore $G$ is a group |
| | **(3)** | |

## Part (c)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $0$ has order $1$ and $1, 2$ and $3$ have order $2$ | M1 | States correctly the order of at least two elements |
| | A1 | States correctly the order of all four elements |
| | **(2)** | |

## Part (d)

| Answer | Marks | Guidance |
|--------|-------|----------|
| There is no element with order $4$ therefore $G$ is not a cyclic group; every element is its own inverse therefore no group generator therefore $G$ is not a cyclic group | B1 | See scheme |
| | **(1)** | |

**Total: 8 marks**
\begin{enumerate}
  \item The operation * is defined on the set $G = \{ 0,1,2,3 \}$ by
\end{enumerate}

$$x ^ { * } y \equiv x + y - 2 x y ( \bmod 4 )$$

(a) Complete the Cayley table below.

\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
* & 0 & 1 & 2 & 3 \\
\hline
0 &  &  &  &  \\
\hline
1 &  &  &  &  \\
\hline
2 &  &  &  &  \\
\hline
3 &  &  &  &  \\
\hline
\end{tabular}
\end{center}

(b) Show that $G$ is a group under the operation *\\
(You may assume the associative law is satisfied.)\\
(c) State the order of each element of $G$.\\
(d) State whether $G$ is a cyclic group, giving a reason for your answer.

\hfill \mbox{\textit{Edexcel FP2 AS 2023 Q1 [8]}}
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