OCR MEI FP3 2014 June — Question 4 24 marks

Exam BoardOCR MEI
ModuleFP3 (Further Pure Mathematics 3)
Year2014
SessionJune
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeSubgroups and cosets
DifficultyChallenging +1.8 This is a substantial Further Maths group theory question requiring multiple techniques: computing element orders, verifying subgroup properties, identifying cyclic subgroups, and working with modular arithmetic groups. While systematic rather than requiring deep insight, the length (6 parts), abstract nature of group theory, and need to work with both abstract and concrete group representations place it well above average difficulty for A-level.
Spec8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups8.03g Cyclic groups: meaning of the term8.03h Generators: of cyclic and non-cyclic groups8.03i Properties of groups: structure of finite groups up to order 7
4 The twelve distinct elements of an abelian multiplicative group \(G\) are $$e , a , a ^ { 2 } , a ^ { 3 } , a ^ { 4 } , a ^ { 5 } , b , a b , a ^ { 2 } b , a ^ { 3 } b , a ^ { 4 } b , a ^ { 5 } b$$ where \(e\) is the identity element, \(a ^ { 6 } = e\) and \(b ^ { 2 } = e\).
  1. Show that the element \(a ^ { 2 } b\) has order 6 .
  2. Show that \(\left\{ e , a ^ { 3 } , b , a ^ { 3 } b \right\}\) is a subgroup of \(G\).
  3. List all the cyclic subgroups of \(G\). You are given that the set $$H = \{ 1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59,61,67,71,73,77,79,83,89 \}$$ with binary operation multiplication modulo 90 is a group.
  4. Determine the order of each of the elements 11, 17 and 19 .
  5. Give a cyclic subgroup of \(H\) with order 4.
  6. By identifying possible values for the elements \(a\) and \(b\) above, or otherwise, give one example of each of the following:
    (A) a non-cyclic subgroup of \(H\) with order 12,
    (B) a non-cyclic subgroup of \(H\) with order 4.
Question 4(i):
AnswerMarks Guidance
AnswerMarks Guidance
\((a^2b)^2 = a^4b^2 = a^4\)M1 Finding one power
\((a^2b)^3 = b,\ (a^2b)^4 = a^2,\ (a^2b)^5 = a^4b\)A1 Three powers correct
\((a^2b)^6 = e\)
Hence \(a^2b\) has order 6E1 Fully correct explanation
[3] No need to state conclusion, provided it has been fully justified
Question 4(ii):
AnswerMarks Guidance
AnswerMarks Guidance
Cayley table for \(\{e, a^3, b, a^3b\}\) (full table shown in mark scheme)B2 Give B1 for no more than three errors or omissions
The set is closed; hence it is a subgroup of \(G\)B1 'Closed' (or equivalent) is required
[3]
Question 4(iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\{e, a^3\}\), \(\{e, b\}\), \(\{e, a^3b\}\)B2 Give B1 for one correct; B0 if any other set of order 3
\(\{e, a^2, a^4\}\)B1
\(\{e, a, a^2, a^3, a^4, a^5\}\)B1 Deduct one mark (out of B3) for each set of order 6 in excess of 3
\(\{e, a^2b, a^4, b, a^2, a^4b\}\)B1 No mark for \(\{e\}\). Deduct one mark (out of B6) for each set (including G) of order other than 1, 2, 3, 6
\(\{e, ab, a^2, a^3b, a^4, a^5b\}\)B1 Deduct one mark (out of B2) for each set of order 2 in excess of 3
[6]
Question 4(iv):
AnswerMarks Guidance
AnswerMarks Guidance
\(11^2=31,\ 11^3=71,\ 11^4=61,\ 11^5=41,\ 11^6=1\)M1 Finding at least two powers of 11 (or 17)
\(17^2=19,\ 17^3=53,\ 17^4=1\)
11 has order 6A1 Either correct implies M1
17 has order 4A1
\(19^2=1\); 19 has order 2B1
[4]
Question 4(v):
AnswerMarks Guidance
AnswerMarks Guidance
\(\{1, 17, 19, 53\}\)M1 Selecting powers of 17
A1Or B2 for \(\{1, 37, 19, 73\}\)
[2]
Question 4(vi)(A):
AnswerMarks Guidance
AnswerMarks Guidance
Taking \(a=11,\ b=19\)B1 There are (many) other possibilities
\(1, 11, 11^2, \ldots, 11^5, 19, 11\times19, 11^2\times19, \ldots, 11^5\times19\)M1 Finding elements of \(G\) using their \(a, b\)
\(\{1, 11, 31, 71, 61, 41, 19, 29, 49, 89, 79, 59\}\)A1
i.e. \(\{1, 11, 19, 29, 31, 41, 49, 59, 61, 71, 79, 89\}\)
[3]
Question 4(vi)(B):
AnswerMarks Guidance
AnswerMarks Guidance
M1Reference to group in (ii)
\(1,\ 11^3,\ 19,\ 11^3\times19\)M1 Finding group in (ii) with their \(a, b\)
\(\{1, 71, 19, 89\}\)A1
[3] 
## Question 4(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(a^2b)^2 = a^4b^2 = a^4$ | M1 | Finding one power |
| $(a^2b)^3 = b,\ (a^2b)^4 = a^2,\ (a^2b)^5 = a^4b$ | A1 | Three powers correct |
| $(a^2b)^6 = e$ | | |
| Hence $a^2b$ has order 6 | E1 | Fully correct explanation |
| **[3]** | | No need to state conclusion, provided it has been fully justified |

## Question 4(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley table for $\{e, a^3, b, a^3b\}$ (full table shown in mark scheme) | B2 | Give B1 for no more than three errors or omissions |
| The set is closed; hence it is a subgroup of $G$ | B1 | 'Closed' (or equivalent) is required |
| **[3]** | | |

## Question 4(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\{e, a^3\}$, $\{e, b\}$, $\{e, a^3b\}$ | B2 | Give B1 for one correct; B0 if any other set of order 3 |
| $\{e, a^2, a^4\}$ | B1 | |
| $\{e, a, a^2, a^3, a^4, a^5\}$ | B1 | Deduct one mark (out of B3) for each set of order 6 in excess of 3 |
| $\{e, a^2b, a^4, b, a^2, a^4b\}$ | B1 | No mark for $\{e\}$. Deduct one mark (out of B6) for each set (including G) of order other than 1, 2, 3, 6 |
| $\{e, ab, a^2, a^3b, a^4, a^5b\}$ | B1 | Deduct one mark (out of B2) for each set of order 2 in excess of 3 |
| **[6]** | | |

## Question 4(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $11^2=31,\ 11^3=71,\ 11^4=61,\ 11^5=41,\ 11^6=1$ | M1 | Finding at least two powers of 11 (or 17) |
| $17^2=19,\ 17^3=53,\ 17^4=1$ | | |
| 11 has order 6 | A1 | Either correct implies M1 |
| 17 has order 4 | A1 | |
| $19^2=1$; 19 has order 2 | B1 | |
| **[4]** | | |

## Question 4(v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\{1, 17, 19, 53\}$ | M1 | Selecting powers of 17 |
| | A1 | Or B2 for $\{1, 37, 19, 73\}$ |
| **[2]** | | |

## Question 4(vi)(A):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking $a=11,\ b=19$ | B1 | There are (many) other possibilities |
| $1, 11, 11^2, \ldots, 11^5, 19, 11\times19, 11^2\times19, \ldots, 11^5\times19$ | M1 | Finding elements of $G$ using their $a, b$ |
| $\{1, 11, 31, 71, 61, 41, 19, 29, 49, 89, 79, 59\}$ | A1 | |
| i.e. $\{1, 11, 19, 29, 31, 41, 49, 59, 61, 71, 79, 89\}$ | | |
| **[3]** | | |

## Question 4(vi)(B):

| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Reference to group in (ii) |
| $1,\ 11^3,\ 19,\ 11^3\times19$ | M1 | Finding group in (ii) with their $a, b$ |
| $\{1, 71, 19, 89\}$ | A1 | |
| **[3]** | | |
4 The twelve distinct elements of an abelian multiplicative group $G$ are

$$e , a , a ^ { 2 } , a ^ { 3 } , a ^ { 4 } , a ^ { 5 } , b , a b , a ^ { 2 } b , a ^ { 3 } b , a ^ { 4 } b , a ^ { 5 } b$$

where $e$ is the identity element, $a ^ { 6 } = e$ and $b ^ { 2 } = e$.
\begin{enumerate}[label=(\roman*)]
\item Show that the element $a ^ { 2 } b$ has order 6 .
\item Show that $\left\{ e , a ^ { 3 } , b , a ^ { 3 } b \right\}$ is a subgroup of $G$.
\item List all the cyclic subgroups of $G$.

You are given that the set

$$H = \{ 1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59,61,67,71,73,77,79,83,89 \}$$

with binary operation multiplication modulo 90 is a group.
\item Determine the order of each of the elements 11, 17 and 19 .
\item Give a cyclic subgroup of $H$ with order 4.
\item By identifying possible values for the elements $a$ and $b$ above, or otherwise, give one example of each of the following:\\
(A) a non-cyclic subgroup of $H$ with order 12,\\
(B) a non-cyclic subgroup of $H$ with order 4.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI FP3 2014 Q4 [24]}}
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