Edexcel FP2 AS 2019 June — Question 4 7 marks

Exam BoardEdexcel
ModuleFP2 AS (Further Pure 2 AS)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeComplete or analyse Cayley table
DifficultyStandard +0.3 This is a guided Cayley table completion with scaffolded steps. Parts (a)(i) and (a)(ii) are shown for students, making the table completion in (b) straightforward. Parts (c) and (d) require only table lookup and applying Lagrange's theorem. While group theory is a Further Maths topic (inherently harder), the question requires minimal problem-solving—mostly following given information and standard procedures. Slightly easier than average A-level difficulty overall.
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 group8.03k Lagrange's theorem: order of subgroup divides order of group
  1. The set \(\{ e , p , q , r , s \}\) forms a group, \(A\), under the operation *
Given that \(e\) is the identity element and that $$p ^ { * } p = s \quad s ^ { * } s = r \quad p ^ { * } p ^ { * } p = q$$
  1. show that
    1. \(p ^ { * } q = r\)
    2. \(s ^ { * } p = q\)
  2. Hence complete the Cayley table below.
    *\(e\)\(\boldsymbol { p }\)\(\boldsymbol { q }\)\(r\)\(s\)
    \(e\)
    \(\boldsymbol { p }\)
    \(\boldsymbol { q }\)
    \(\boldsymbol { r }\)
    \(S\)
    A spare table can be found on page 11 if you need to rewrite your Cayley table.
  3. Use your table to find \(p ^ { * } q ^ { * } r ^ { * } s\) A student states that there is a subgroup of \(A\) of order 3
  4. Comment on the validity of this statement, giving a reason for your answer. \includegraphics[max width=\textwidth, alt={}, center]{989d779e-c40a-4658-ad98-17a37ab1d9e1-11_2464_74_304_36}
    Only use this grid if you need to rewrite the Cayley table.
    *\(e\)\(\boldsymbol { p }\)\(\boldsymbol { q }\)\(r\)\(s\)
    \(e\)
    \(\boldsymbol { p }\)
    \(\boldsymbol { q }\)
    \(\boldsymbol { r }\)
    \(S\)
Question 4:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(p*q = p*p*p*p = s*s = r\); OR \(s*s=r\Rightarrow p*p*p*p=r\Rightarrow p*q=r\)B1 Correct proof for printed statement
\(s*p = p*p*p = q\); OR as \(p*p*p=q\) and \(p*p=s\Rightarrow s*p=q\)B1 Correct proof for printed statement
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Cayley table with at least 13 correct entries (highlighted entries)M1 Finds at least 13 correct entries
Completely correct tableA1 Fully correct table
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(p*q*r*s = e\)B1 See scheme
Part (d)
AnswerMarks Guidance
Answer/WorkingMark Guidance
The order of a subgroup is a factor of the order of the group (Lagrange's Theorem)M1 Some indication that order of subgroup must be factor of order of group; may say 3 is not a factor of 5
As 3 is not a factor of 5, the student's statement is wrongA1 Fully correct unambiguous statement referencing Lagrange's theorem and that 3 does not divide 5; no contradictory statements 
## Question 4:

**Part (a)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $p*q = p*p*p*p = s*s = r$; OR $s*s=r\Rightarrow p*p*p*p=r\Rightarrow p*q=r$ | B1 | Correct proof for printed statement |
| $s*p = p*p*p = q$; OR as $p*p*p=q$ and $p*p=s\Rightarrow s*p=q$ | B1 | Correct proof for printed statement |

**Part (b)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Cayley table with at least 13 correct entries (highlighted entries) | M1 | Finds at least 13 correct entries |
| Completely correct table | A1 | Fully correct table |

**Part (c)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $p*q*r*s = e$ | B1 | See scheme |

**Part (d)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| The order of a subgroup is a factor of the order of the group (Lagrange's Theorem) | M1 | Some indication that order of subgroup must be factor of order of group; may say 3 is not a factor of 5 |
| As 3 is not a factor of 5, the student's statement is wrong | A1 | Fully correct unambiguous statement referencing Lagrange's theorem and that 3 does not divide 5; no contradictory statements |
\begin{enumerate}
  \item The set $\{ e , p , q , r , s \}$ forms a group, $A$, under the operation *
\end{enumerate}

Given that $e$ is the identity element and that

$$p ^ { * } p = s \quad s ^ { * } s = r \quad p ^ { * } p ^ { * } p = q$$

(a) show that\\
(i) $p ^ { * } q = r$\\
(ii) $s ^ { * } p = q$\\
(b) Hence complete the Cayley table below.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
* & $e$ & $\boldsymbol { p }$ & $\boldsymbol { q }$ & $r$ & $s$ \\
\hline
$e$ &  &  &  &  &  \\
\hline
$\boldsymbol { p }$ &  &  &  &  &  \\
\hline
$\boldsymbol { q }$ &  &  &  &  &  \\
\hline
$\boldsymbol { r }$ &  &  &  &  &  \\
\hline
$S$ &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}

A spare table can be found on page 11 if you need to rewrite your Cayley table.\\
(c) Use your table to find $p ^ { * } q ^ { * } r ^ { * } s$

A student states that there is a subgroup of $A$ of order 3\\
(d) Comment on the validity of this statement, giving a reason for your answer.

\includegraphics[max width=\textwidth, alt={}, center]{989d779e-c40a-4658-ad98-17a37ab1d9e1-11_2464_74_304_36}\\

Only use this grid if you need to rewrite the Cayley table.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
* & $e$ & $\boldsymbol { p }$ & $\boldsymbol { q }$ & $r$ & $s$ \\
\hline
$e$ &  &  &  &  &  \\
\hline
$\boldsymbol { p }$ &  &  &  &  &  \\
\hline
$\boldsymbol { q }$ &  &  &  &  &  \\
\hline
$\boldsymbol { r }$ &  &  &  &  &  \\
\hline
$S$ &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}

\hfill \mbox{\textit{Edexcel FP2 AS 2019 Q4 [7]}}
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