Question 4 - A-Level Maths

Edexcel FP2 AS Specimen — Question 4 11 marks

Exam Board	Edexcel
Module	FP2 AS (Further Pure 2 AS)
Session	Specimen
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Groups
Type	Verify group axioms
Difficulty	Standard +0.3 This is a routine Further Maths group theory question requiring completion of a Cayley table using modular arithmetic, verification of standard group axioms (closure, identity, inverses), finding element order, and identifying a generator. While it's a Further Maths topic, the question follows a standard template with straightforward calculations and no novel problem-solving required.
Spec	8.03a Binary operations: and their properties on given sets 8.03b Cayley tables: construct for finite sets under binary operation 8.03c Group definition: recall and use, show structure is/isn't a group 8.03e Order of elements: and order of groups 8.03g Cyclic groups: meaning of the term 8.03h Generators: of cyclic and non-cyclic groups

The operation * is defined on the set \(S = \{ 0,2,3,4,5,6 \}\) by \(x ^ { * } y = x + y = x y ( \bmod 7 )\)

*	0	2	3	4	5	6
0
2		0
3						5
4
5		4
6

1. Complete the Cayley table shown above
2. Show that \(S\) is a group under the operation *
  (You may assume the associative law is satisfied.)
Show that the element 4 has order 3
Find an element which generates the group and express each of the elements in terms of this generator.

Show mark scheme Show mark scheme source

Question 4:

Part (a)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Begins completing the table — correct first row and first column using symmetry	M1	Begins completing the table using symmetry
Mostly correct — three rows or three columns correct (demonstrating understanding of using \(*\))	M1	Mostly correct
Completely correct table	A1	Completely correct

Part (a)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Identity is zero and there is closure as shown above	M1	States closure and identifies the identity as zero
3 and 5 are inverses, 4 and 6 are inverses, 2 is self-inverse, 0 is identity so is self-inverse	M1	Finds inverses for each element
Associative law may be assumed so \(S\) forms a group	A1	States that associative law is satisfied and so all axioms satisfied and \(S\) is a group

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(444=4(44)=46\) or \(444=(44)4=64\)	M1	Clearly begins process to find \(444\) reaching \(64\) or \(46\) with clear explanation
\(=0\) (the identity) so 4 has order 3	A1	Gives answer as zero, states identity and deduces that order is 3

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
3 and 5 each have order 6 so either generates the group	M1	Finds either 3 or 5 or both
Either \(3^1=3,\ 3^2=4,\ 3^3=2,\ 3^4=6,\ 3^5=5,\ 3^6=0\)	A1	Expresses four of the six terms as powers of either generator correctly (may omit identity and generator itself)
Or \(5^1=5,\ 5^2=6,\ 5^3=2,\ 5^4=4,\ 5^5=3,\ 5^6=0\)	A1	Expresses all six terms correctly in terms of either 3 or 5 (do not need to give both)

## Question 4:

### Part (a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Begins completing the table — correct first row and first column using symmetry | M1 | Begins completing the table using symmetry |
| Mostly correct — three rows or three columns correct (demonstrating understanding of using $*$) | M1 | Mostly correct |
| Completely correct table | A1 | Completely correct |

### Part (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Identity is zero and there is closure as shown above | M1 | States closure and identifies the identity as zero |
| 3 and 5 are inverses, 4 and 6 are inverses, 2 is self-inverse, 0 is identity so is self-inverse | M1 | Finds inverses for each element |
| Associative law may be assumed so $S$ forms a group | A1 | States that associative law is satisfied and so all axioms satisfied and $S$ is a group |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4*4*4=4*(4*4)=4*6$ or $4*4*4=(4*4)*4=6*4$ | M1 | Clearly begins process to find $4*4*4$ reaching $6*4$ or $4*6$ with clear explanation |
| $=0$ (the identity) so 4 has order 3 | A1 | Gives answer as zero, states identity and deduces that order is 3 |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| 3 and 5 each have order 6 so either generates the group | M1 | Finds either 3 or 5 or both |
| Either $3^1=3,\ 3^2=4,\ 3^3=2,\ 3^4=6,\ 3^5=5,\ 3^6=0$ | A1 | Expresses four of the six terms as powers of either generator correctly (may omit identity and generator itself) |
| Or $5^1=5,\ 5^2=6,\ 5^3=2,\ 5^4=4,\ 5^5=3,\ 5^6=0$ | A1 | Expresses all six terms correctly in terms of either 3 or 5 (do not need to give both) |

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Show LaTeX source

\begin{enumerate}
  \item The operation * is defined on the set $S = \{ 0,2,3,4,5,6 \}$ by $x ^ { * } y = x + y = x y ( \bmod 7 )$
\end{enumerate}

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
* & 0 & 2 & 3 & 4 & 5 & 6 \\
\hline
0 &  &  &  &  &  &  \\
\hline
2 &  & 0 &  &  &  &  \\
\hline
3 &  &  &  &  &  & 5 \\
\hline
4 &  &  &  &  &  &  \\
\hline
5 &  & 4 &  &  &  &  \\
\hline
6 &  &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}

(a) (i) Complete the Cayley table shown above\\
(ii) Show that $S$ is a group under the operation *\\
(You may assume the associative law is satisfied.)\\
(b) Show that the element 4 has order 3\\
(c) Find an element which generates the group and express each of the elements in terms of this generator.

\hfill \mbox{\textit{Edexcel FP2 AS  Q4 [11]}}

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