Question 8 - A-Level Maths

OCR FP3 2016 June — Question 8 17 marks

Exam Board	OCR
Module	FP3 (Further Pure Mathematics 3)
Year	2016
Session	June
Marks	17
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Groups
Type	Subgroups and cosets
Difficulty	Challenging +1.8 This is a Further Maths group theory question requiring understanding of non-commutative groups, proof by contradiction, and systematic construction of subgroups. While the individual steps are guided, it demands abstract algebraic reasoning beyond standard A-level, including deducing group relations and constructing Cayley tables for the dihedral group D_4.
Spec	8.03c Group definition: recall and use, show structure is/isn't a group 8.03d Latin square property: for group tables 8.03e Order of elements: and order of groups 8.03f Subgroups: definition and tests for proper subgroups

8 A non-commutative multiplicative group $G$ of order eight has the elements $$\left\{ e , a , a ^ { 2 } , a ^ { 3 } , b , a b , a ^ { 2 } b , a ^ { 3 } b \right\}$$ where $e$ is the identity and $a ^ { 4 } = b ^ { 2 } = e$.

Show that $b a \neq a ^ { n }$ for any integer $n$.
Prove, by contradiction, that $b a \neq a ^ { 2 } b$ and also that $b a \neq a b$. Deduce that $b a = a ^ { 3 } b$.
Prove that $b a ^ { 2 } = a ^ { 2 } b$.
Construct group tables for the three subgroups of $G$ of order four. \section*{END OF QUESTION PAPER}

Show mark scheme Show mark scheme source

Question 8:

Part (i):

Answer	Marks	Guidance
Answer	Marks	Guidance
$ba = a^n \Rightarrow b = a^{n-1}$	M1
But these are distinct elements so $ba \neq a^n$	A1

Part (ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
$ba = a^2b$ $\Rightarrow a^2ba = a^4b$ $\Rightarrow a^2ba = b$	M1	or $b = a^2ba^3$, $a = ba^2b$, $a^2 = bab$ or $b = ba^2$
$\Rightarrow a^2ba^4 = ba^3$ $\Rightarrow a^2b = ba^3$ $\Rightarrow ba = ba^3$ $\Rightarrow e = a^2$	M1	validly reach any equality which gives 2 distinct elements of the group as equal
Which is false, hence $ba \neq a^2b$	A1	Complete argument
If $ba = ab$ then (all element pairs would have to be commutative and so) $G$ would be abelian.	M1	Do not award for $G$ non-abelian $\Rightarrow ba \neq ab$
If $ba = b$ then $a = e$ so $ba \neq b$	M1
So, by elimination of other possibilities, $ba = a^3b$	A1	Dependent on all previous marks

Part (iii):

Answer	Marks	Guidance
Answer	Marks	Guidance
$ba^2 = baa = a^3ba$	M1	Use previous result
$= a^3 \cdot a^3b = a^2b$	A1	Complete argument

Part (iv):

Answer	Marks	Guidance
Answer	Marks	Guidance
Cayley table for $\{e, a, a^2, a^3\}$: rows/columns headed $e, a, a^2, a^3$ with entries $e,a,a^2,a^3$ / $a,a^2,a^3,e$ / $a^2,a^3,e,a$ / $a^3,e,a,a^2$	B1	All correct
Cayley table for $\{e, a^2, b, a^2b\}$: correct header elements	B1	Correct elements
At least 12 out of 16 entries correct	M1
All entries correct	A1
Cayley table for $\{e, a^2, ab, a^3b\}$: correct header elements	B1	Correct elements
At least 12 out of 16 entries correct	M1
All entries correct	A1

# Question 8:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $ba = a^n \Rightarrow b = a^{n-1}$ | M1 | |
| But these are distinct elements so $ba \neq a^n$ | A1 | |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $ba = a^2b$ $\Rightarrow a^2ba = a^4b$ $\Rightarrow a^2ba = b$ | M1 | or $b = a^2ba^3$, $a = ba^2b$, $a^2 = bab$ or $b = ba^2$ |
| $\Rightarrow a^2ba^4 = ba^3$ $\Rightarrow a^2b = ba^3$ $\Rightarrow ba = ba^3$ $\Rightarrow e = a^2$ | M1 | validly reach any equality which gives 2 distinct elements of the group as equal |
| Which is false, hence $ba \neq a^2b$ | A1 | Complete argument |
| If $ba = ab$ then (all element pairs would have to be commutative and so) $G$ would be abelian. | M1 | Do not award for $G$ non-abelian $\Rightarrow ba \neq ab$ |
| If $ba = b$ then $a = e$ so $ba \neq b$ | M1 | |
| So, by elimination of other possibilities, $ba = a^3b$ | A1 | Dependent on all previous marks |

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $ba^2 = baa = a^3ba$ | M1 | Use previous result |
| $= a^3 \cdot a^3b = a^2b$ | A1 | Complete argument |

## Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley table for $\{e, a, a^2, a^3\}$: rows/columns headed $e, a, a^2, a^3$ with entries $e,a,a^2,a^3$ / $a,a^2,a^3,e$ / $a^2,a^3,e,a$ / $a^3,e,a,a^2$ | B1 | All correct |
| Cayley table for $\{e, a^2, b, a^2b\}$: correct header elements | B1 | Correct elements |
| At least 12 out of 16 entries correct | M1 | |
| All entries correct | A1 | |
| Cayley table for $\{e, a^2, ab, a^3b\}$: correct header elements | B1 | Correct elements |
| At least 12 out of 16 entries correct | M1 | |
| All entries correct | A1 | |

Show LaTeX source

8 A non-commutative multiplicative group $G$ of order eight has the elements

$$\left\{ e , a , a ^ { 2 } , a ^ { 3 } , b , a b , a ^ { 2 } b , a ^ { 3 } b \right\}$$

where $e$ is the identity and $a ^ { 4 } = b ^ { 2 } = e$.\\
(i) Show that $b a \neq a ^ { n }$ for any integer $n$.\\
(ii) Prove, by contradiction, that $b a \neq a ^ { 2 } b$ and also that $b a \neq a b$. Deduce that $b a = a ^ { 3 } b$.\\
(iii) Prove that $b a ^ { 2 } = a ^ { 2 } b$.\\
(iv) Construct group tables for the three subgroups of $G$ of order four.

\section*{END OF QUESTION PAPER}

\hfill \mbox{\textit{OCR FP3 2016 Q8 [17]}}

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