Pre-U Pre-U 9795/1 2019 Specimen — Question 8 5 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2019
SessionSpecimen
Marks5
TopicGroups
TypeMatrix groups
DifficultyChallenging +1.8 This is a challenging group theory question requiring proof of group axioms for a non-standard matrix set, identification of subgroups, and application of Lagrange's theorem. While the matrix form is simple, students must verify closure (non-trivial calculation showing the form is preserved), find the identity, prove inverses exist, then systematically find subgroups of specific orders. This requires deep understanding of abstract algebra concepts beyond routine application, placing it well above average difficulty but not at the extreme end since the calculations themselves are manageable.
Spec4.03b Matrix operations: addition, multiplication, scalar8.03c Group definition: recall and use, show structure is/isn't a group8.03g Cyclic groups: meaning of the term

8 Consider the set \(S\) of all matrices of the form \(\left( \begin{array} { l l } p & p \\ p & p \end{array} \right)\), where p is a non-zero rational number.
  1. Show that \(S\), under the operation of matrix multiplication, forms a group, \(G\). (You may assume that matrix multiplication is associative.)
  2. Find a subgroup of \(G\) of order 2 and show that \(G\) contains no subgroups of order 3 .

(a) Good attempt to multiply 2 matrices of the appropriate form:
\(\begin{pmatrix}p & p\\p & p\end{pmatrix}\begin{pmatrix}q & q\\q & q\end{pmatrix}\) M1
"Closure" noted or implied by correct product matrix \(= \begin{pmatrix}2pq & 2pq\\2pq & 2pq\end{pmatrix} \in S\) A1
Statement that \(\times_M\) known to be associative OR Alternative \([(p)(q)](r) = (p)[(q)(r)] = (4pqr)\) shown B1
Identity is \(\begin{pmatrix}\frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2}\end{pmatrix}\) \((\in S)\) B1
\(\begin{pmatrix}p & p\\p & p\end{pmatrix}^{-1} = \begin{pmatrix}\frac{1}{4p} & \frac{1}{4p}\\\frac{1}{4p} & \frac{1}{4p}\end{pmatrix}\) \((\in S\) as \(p \uparrow 0)\) B1
\(\ldots\) and \((S, \times_M)\) is a group since all four group axioms are satisfied
Total: 5
(b) Attempt to look for a self-inverse element; i.e. solving \(p = \frac{1}{4p}\) using their \((p)^{-1}\) and E M1
\(p = -\frac{1}{2}\) and noting that \(H = \{\mathbf{E}, \mathbf{A}\}\) where \(\mathbf{E} = \begin{pmatrix}\frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2}\end{pmatrix}\), \(\mathbf{A} = \begin{pmatrix}-\frac{1}{2} & -\frac{1}{2}\\-\frac{1}{2} & -\frac{1}{2}\end{pmatrix}\) A1
Looking for \(\{\mathbf{E}, \mathbf{B}, \mathbf{B}^2\}\) where \(\mathbf{B}^3 = \mathbf{E}\); i.e. solving \((4p^3) = \frac{1}{2}\) M1
Explaining carefully that \(p^3 = \frac{1}{8} \Leftrightarrow p = \frac{1}{2}\) and no such \(\mathbf{B}\) \((\neq \mathbf{E})\) exists A1
Total: 4
**(a)** Good attempt to multiply 2 matrices of the appropriate form:
$\begin{pmatrix}p & p\\p & p\end{pmatrix}\begin{pmatrix}q & q\\q & q\end{pmatrix}$ **M1**

"Closure" noted or implied by correct product matrix $= \begin{pmatrix}2pq & 2pq\\2pq & 2pq\end{pmatrix} \in S$ **A1**

Statement that $\times_M$ known to be associative **OR** Alternative $[(p)(q)](r) = (p)[(q)(r)] = (4pqr)$ shown **B1**

Identity is $\begin{pmatrix}\frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2}\end{pmatrix}$ $(\in S)$ **B1**

$\begin{pmatrix}p & p\\p & p\end{pmatrix}^{-1} = \begin{pmatrix}\frac{1}{4p} & \frac{1}{4p}\\\frac{1}{4p} & \frac{1}{4p}\end{pmatrix}$ $(\in S$ as $p \uparrow 0)$ **B1**

$\ldots$ and $(S, \times_M)$ is a group since all four group axioms are satisfied

**Total: 5**

**(b)** Attempt to look for a self-inverse element; i.e. solving $p = \frac{1}{4p}$ using their $(p)^{-1}$ and **E** **M1**

$p = -\frac{1}{2}$ and noting that $H = \{\mathbf{E}, \mathbf{A}\}$ where $\mathbf{E} = \begin{pmatrix}\frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2}\end{pmatrix}$, $\mathbf{A} = \begin{pmatrix}-\frac{1}{2} & -\frac{1}{2}\\-\frac{1}{2} & -\frac{1}{2}\end{pmatrix}$ **A1**

Looking for $\{\mathbf{E}, \mathbf{B}, \mathbf{B}^2\}$ where $\mathbf{B}^3 = \mathbf{E}$; i.e. solving $(4p^3) = \frac{1}{2}$ **M1**

Explaining carefully that $p^3 = \frac{1}{8} \Leftrightarrow p = \frac{1}{2}$ and no such $\mathbf{B}$ $(\neq \mathbf{E})$ exists **A1**

**Total: 4**
8 Consider the set $S$ of all matrices of the form $\left( \begin{array} { l l } p & p \\ p & p \end{array} \right)$, where p is a non-zero rational number.
\begin{enumerate}[label=(\alph*)]
\item Show that $S$, under the operation of matrix multiplication, forms a group, $G$. (You may assume that matrix multiplication is associative.)
\item Find a subgroup of $G$ of order 2 and show that $G$ contains no subgroups of order 3 .
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2019 Q8 [5]}}