Question 4 - A-Level Maths

OCR MEI FP3 2010 June — Question 4 24 marks

Exam Board	OCR MEI
Module	FP3 (Further Pure Mathematics 3)
Year	2010
Session	June
Marks	24
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Groups
Type	Function composition groups
Difficulty	Challenging +1.8 This is a substantial Further Maths question on abstract group theory requiring multiple techniques: function composition verification, completing a Cayley table, finding inverses and subgroups, determining element orders in complex number groups, proving a group is cyclic, and analyzing isomorphisms. While each individual part uses standard methods, the length (8 parts), conceptual sophistication (abstract algebra), and requirement to connect three different group structures places this well above average A-level difficulty but below the most challenging proof-based questions.
Spec	8.03c Group definition: recall and use, show structure is/isn't a group 8.03e Order of elements: and order of groups 8.03f Subgroups: definition and tests for proper subgroups 8.03g Cyclic groups: meaning of the term

4 The group $F = \{ \mathrm { p } , \mathrm { q } , \mathrm { r } , \mathrm { s } , \mathrm { t } , \mathrm { u } \}$ consists of the six functions defined by $$\mathrm { p } ( x ) = x \quad \mathrm { q } ( x ) = 1 - x \quad \mathrm { r } ( x ) = \frac { 1 } { x } \quad \mathrm {~s} ( x ) = \frac { x - 1 } { x } \quad \mathrm { t } ( x ) = \frac { x } { x - 1 } \quad \mathrm { u } ( x ) = \frac { 1 } { 1 - x } ,$$ the binary operation being composition of functions.

Show that st $= \mathrm { r }$ and find ts.
Copy and complete the following composition table for $F$.
p q r s t u
p p q r s t u
q q p s r u t
r r u p t s q
s s t q u r p
t t s u
u u r t
Give the inverse of each element of $F$.
List all the subgroups of $F$. The group $M$ consists of $\left\{ 1 , - 1 , e ^ { \frac { \pi } { 3 } \mathrm { j } } , e ^ { - \frac { \pi } { 3 } \mathrm { j } } , e ^ { \frac { 2 \pi } { 3 } \mathrm { j } } , e ^ { - \frac { 2 \pi } { 3 } \mathrm { j } } \right\}$ with multiplication of complex numbers as its binary operation.
Find the order of each element of $M$. The group $G$ consists of the positive integers between 1 and 18 inclusive, under multiplication modulo 19.
Show that $G$ is a cyclic group which can be generated by the element 2 .
Explain why $G$ has no subgroup which is isomorphic to $F$.
Find a subgroup of $G$ which is isomorphic to $M$.

Show mark scheme Show mark scheme source

Question 4:

Part (i):

Answer	Marks	Guidance
Answer	Marks	Guidance
$st(x)=s\!\left(\frac{x}{x-1}\right)=\frac{\frac{x}{x-1}-1}{\frac{x}{x-1}}=\frac{x-(x-1)}{x}=\frac{1}{x}=r(x)$	M1, A1 (ag)
$ts(x)=t\!\left(\frac{x-1}{x}\right)=\frac{\frac{x-1}{x}}{\frac{x-1}{x}-1}=\frac{x-1}{(x-1)-x}=1-x=q(x)$	M1, A1	[4]

Part (ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Cayley table completed correctly	B3	Give B2 for 4 correct, B1 for 2 correct [3]

Part (iii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Inverses: $p\to p$, $q\to r$, $r\to q$... (table shown)	B3	Give B2 for 4 correct, B1 for 2 correct [3]

Part (iv):

Answer	Marks	Guidance
Answer	Marks	Guidance
$\{p\}$, $F$	B1B1B1	Ignore these in the marking
$\{p,q\}$, $\{p,r\}$, $\{p,t\}$	B1	Deduct one mark for each non-trivial subgroup in excess of four [4]
$\{p,s,u\}$

Part (v):

Answer	Marks	Guidance
Answer	Marks	Guidance
Orders: $1\to1$, $-1\to2$, $e^{\frac{\pi}{3}j}\to6$, $e^{-\frac{\pi}{3}j}\to6$, $e^{\frac{2\pi}{3}j}\to3$, $e^{-\frac{2\pi}{3}j}\to3$	B4	Give B3 for 4 correct, B2 for 3 correct, B1 for 2 correct [4]

Part (vi):

Answer	Marks	Guidance
Answer	Marks	Guidance
$2^1=2,\ 2^2=4,\ 2^3=8,\ 2^4=16,\ 2^5=13,\ 2^6=7$	M1	Finding (at least two) powers of 2
$2^7=14,\ 2^8=9,\ 2^9=18,\ 2^{10}=17,\ 2^{11}=15,\ 2^{12}=11$	A1	For $2^6=7$ and $2^9=18$
$2^{13}=3,\ 2^{14}=6,\ 2^{15}=12,\ 2^{16}=5,\ 2^{17}=10,\ 2^{18}=1$
Hence 2 has order 18	A1	Correctly shown; all powers listed implies final A1 [3]

Part (vii):

Answer	Marks	Guidance
Answer	Marks	Guidance
$G$ is abelian (so all its subgroups are abelian); $F$ is not abelian	B1	Can have 'cyclic' instead of 'abelian' [1]

Part (viii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Subgroup of order 6 is $\{1, 2^3, 2^6, 2^9, 2^{12}, 2^{15}\}$ i.e. $\{1,7,8,11,12,18\}$	M1, A1	or B2 [2]

# Question 4:

## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $st(x)=s\!\left(\frac{x}{x-1}\right)=\frac{\frac{x}{x-1}-1}{\frac{x}{x-1}}=\frac{x-(x-1)}{x}=\frac{1}{x}=r(x)$ | M1, A1 (ag) | |
| $ts(x)=t\!\left(\frac{x-1}{x}\right)=\frac{\frac{x-1}{x}}{\frac{x-1}{x}-1}=\frac{x-1}{(x-1)-x}=1-x=q(x)$ | M1, A1 | **[4]** |

## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley table completed correctly | B3 | Give B2 for 4 correct, B1 for 2 correct **[3]** |

## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Inverses: $p\to p$, $q\to r$, $r\to q$... (table shown) | B3 | Give B2 for 4 correct, B1 for 2 correct **[3]** |

## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\{p\}$, $F$ | B1B1B1 | Ignore these in the marking |
| $\{p,q\}$, $\{p,r\}$, $\{p,t\}$ | B1 | Deduct one mark for each non-trivial subgroup in excess of four **[4]** |
| $\{p,s,u\}$ | | |

## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Orders: $1\to1$, $-1\to2$, $e^{\frac{\pi}{3}j}\to6$, $e^{-\frac{\pi}{3}j}\to6$, $e^{\frac{2\pi}{3}j}\to3$, $e^{-\frac{2\pi}{3}j}\to3$ | B4 | Give B3 for 4 correct, B2 for 3 correct, B1 for 2 correct **[4]** |

## Part (vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2^1=2,\ 2^2=4,\ 2^3=8,\ 2^4=16,\ 2^5=13,\ 2^6=7$ | M1 | Finding (at least two) powers of 2 |
| $2^7=14,\ 2^8=9,\ 2^9=18,\ 2^{10}=17,\ 2^{11}=15,\ 2^{12}=11$ | A1 | For $2^6=7$ and $2^9=18$ |
| $2^{13}=3,\ 2^{14}=6,\ 2^{15}=12,\ 2^{16}=5,\ 2^{17}=10,\ 2^{18}=1$ | | |
| Hence 2 has order 18 | A1 | Correctly shown; all powers listed implies final A1 **[3]** |

## Part (vii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G$ is abelian (so all its subgroups are abelian); $F$ is not abelian | B1 | Can have 'cyclic' instead of 'abelian' **[1]** |

## Part (viii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Subgroup of order 6 is $\{1, 2^3, 2^6, 2^9, 2^{12}, 2^{15}\}$ i.e. $\{1,7,8,11,12,18\}$ | M1, A1 | or B2 **[2]** |

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Show LaTeX source

4 The group $F = \{ \mathrm { p } , \mathrm { q } , \mathrm { r } , \mathrm { s } , \mathrm { t } , \mathrm { u } \}$ consists of the six functions defined by

$$\mathrm { p } ( x ) = x \quad \mathrm { q } ( x ) = 1 - x \quad \mathrm { r } ( x ) = \frac { 1 } { x } \quad \mathrm {~s} ( x ) = \frac { x - 1 } { x } \quad \mathrm { t } ( x ) = \frac { x } { x - 1 } \quad \mathrm { u } ( x ) = \frac { 1 } { 1 - x } ,$$

the binary operation being composition of functions.\\
(i) Show that st $= \mathrm { r }$ and find ts.\\
(ii) Copy and complete the following composition table for $F$.

\begin{center}
\begin{tabular}{ c | c c c c c c }
 & p & q & r & s & t & u \\
\hline
p & p & q & r & s & t & u \\
q & q & p & s & r & u & t \\
r & r & u & p & t & s & q \\
s & s & t & q & u & r & p \\
t & t & s & u &  &  &  \\
u & u & r & t &  &  &  \\
\end{tabular}
\end{center}

(iii) Give the inverse of each element of $F$.\\
(iv) List all the subgroups of $F$.

The group $M$ consists of $\left\{ 1 , - 1 , e ^ { \frac { \pi } { 3 } \mathrm { j } } , e ^ { - \frac { \pi } { 3 } \mathrm { j } } , e ^ { \frac { 2 \pi } { 3 } \mathrm { j } } , e ^ { - \frac { 2 \pi } { 3 } \mathrm { j } } \right\}$ with multiplication of complex numbers as its binary operation.\\
(v) Find the order of each element of $M$.

The group $G$ consists of the positive integers between 1 and 18 inclusive, under multiplication modulo 19.\\
(vi) Show that $G$ is a cyclic group which can be generated by the element 2 .\\
(vii) Explain why $G$ has no subgroup which is isomorphic to $F$.\\
(viii) Find a subgroup of $G$ which is isomorphic to $M$.

\hfill \mbox{\textit{OCR MEI FP3 2010 Q4 [24]}}

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