Pre-U Pre-U 9795/1 Specimen — Question 6 9 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
SessionSpecimen
Marks9
TopicGroups
TypeVerify group axioms
DifficultyStandard +0.8 This is a comprehensive group theory verification question requiring proof of closure (using a given identity), associativity (algebraic manipulation), finding identity and inverse elements, and identifying a subgroup. While systematic, it demands careful algebraic work across multiple parts and understanding of abstract group axioms, placing it moderately above average difficulty for Further Maths students.
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 group8.03f Subgroups: definition and tests for proper subgroups8.03g Cyclic groups: meaning of the term

6 The set \(S\) consists of all real numbers except 1. The binary operation * is defined for all \(a , b\) in \(S\) by $$a * b = a + b - a b$$
  1. By considering the identity \(a + b - a b \equiv 1 - ( a - 1 ) ( b - 1 )\), or otherwise, show that \(S\) is closed under *.
  2. Show that * is associative on \(S\).
  3. Find the identity of \(S\) under \(*\), and the inverse of \(x\) for all \(x \in S\).
  4. The set \(S\), together with the binary operation *, forms a group \(G\). Find a subgroup of \(G\) of order 2 .

(i) \(a * b = 1 \Leftrightarrow (a-1)(b-1) = 0 \Leftrightarrow a\) or \(b = 1\)
(hence \(a \neq 1,\ b \neq 1 \Rightarrow a*b \neq 1\)): M1, A1 [2]
(ii) \((a*b)*c = (a+b-ab)*c = a+b+c-(ab+bc+ca)+abc\)
\(a*(b*c) = a*(b+c-bc) = a+b+c-(ab+bc+ca)+abc\)
For attempt at both: M1; All correct: A1 [2]
(iii) \(e + x - ex = x \Rightarrow e = 0\): B1
\(y + x - yx = 0 \Rightarrow y = \frac{x}{x-1}\), for attempt using their \(e\): M1
Correct (no need to observe that \(x \neq 1\)): A1 [3]
(iv) \(x = \frac{x}{x-1} \Rightarrow x = 2\) so subgroup is \(\{0, 2\}\)
For attempt at self-inverse element or via \(x*x = 0 \Rightarrow 2x - x^2 = 0\): M1; cao: A1 [2]
**(i)** $a * b = 1 \Leftrightarrow (a-1)(b-1) = 0 \Leftrightarrow a$ or $b = 1$

(hence $a \neq 1,\ b \neq 1 \Rightarrow a*b \neq 1$): M1, A1 **[2]**

**(ii)** $(a*b)*c = (a+b-ab)*c = a+b+c-(ab+bc+ca)+abc$

$a*(b*c) = a*(b+c-bc) = a+b+c-(ab+bc+ca)+abc$

For attempt at both: M1; All correct: A1 **[2]**

**(iii)** $e + x - ex = x \Rightarrow e = 0$: B1

$y + x - yx = 0 \Rightarrow y = \frac{x}{x-1}$, for attempt using their $e$: M1

Correct (no need to observe that $x \neq 1$): A1 **[3]**

**(iv)** $x = \frac{x}{x-1} \Rightarrow x = 2$ so subgroup is $\{0, 2\}$

For attempt at self-inverse element or via $x*x = 0 \Rightarrow 2x - x^2 = 0$: M1; **cao**: A1 **[2]**
6 The set $S$ consists of all real numbers except 1. The binary operation * is defined for all $a , b$ in $S$ by

$$a * b = a + b - a b$$

(i) By considering the identity $a + b - a b \equiv 1 - ( a - 1 ) ( b - 1 )$, or otherwise, show that $S$ is closed under *.\\
(ii) Show that * is associative on $S$.\\
(iii) Find the identity of $S$ under $*$, and the inverse of $x$ for all $x \in S$.\\
(iv) The set $S$, together with the binary operation *, forms a group $G$. Find a subgroup of $G$ of order 2 .

\hfill \mbox{\textit{Pre-U Pre-U 9795/1  Q6 [9]}}