OCR MEI FP3 2009 June — Question 4 24 marks

Exam BoardOCR MEI
ModuleFP3 (Further Pure Mathematics 3)
Year2009
SessionJune
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeIsomorphism between groups
DifficultyChallenging +1.2 This is a substantial Further Maths group theory question covering multiple concepts (cyclic groups, isomorphisms, subgroups, permutation groups), but each part follows standard procedures. Parts (i)-(iii) are routine verification tasks for cyclic groups and finding generators. Parts (iv)-(vii) involve composition of permutations and identifying S₃ structure, which are textbook exercises for students who have studied group theory. The length and breadth make it moderately challenging, but no part requires novel insight beyond applying learned techniques.
Spec8.03c 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 term8.03h Generators: of cyclic and non-cyclic groups8.03i Properties of groups: structure of finite groups up to order 78.03l Isomorphism: determine using informal methods
4 The group \(G = \{ 1,2,3,4,5,6 \}\) has multiplication modulo 7 as its operation. The group \(H = \{ 1,5,7,11,13,17 \}\) has multiplication modulo 18 as its operation.
  1. Show that the groups \(G\) and \(H\) are both cyclic.
  2. List all the proper subgroups of \(G\).
  3. Specify an isomorphism between \(G\) and \(H\). The group \(S = \{ \mathrm { a } , \mathrm { b } , \mathrm { c } , \mathrm { d } , \mathrm { e } , \mathrm { f } \}\) consists of functions with domain \(\{ 1,2,3 \}\) given by $$\begin{array} { l l l } \mathrm { a } ( 1 ) = 2 & \mathrm { a } ( 2 ) = 3 & \mathrm { a } ( 3 ) = 1 \\ \mathrm {~b} ( 1 ) = 3 & \mathrm {~b} ( 2 ) = 1 & \mathrm {~b} ( 3 ) = 2 \\ \mathrm { c } ( 1 ) = 1 & \mathrm { c } ( 2 ) = 3 & \mathrm { c } ( 3 ) = 2 \\ \mathrm {~d} ( 1 ) = 3 & \mathrm {~d} ( 2 ) = 2 & \mathrm {~d} ( 3 ) = 1 \\ \mathrm { e } ( 1 ) = 1 & \mathrm { e } ( 2 ) = 2 & \mathrm { e } ( 3 ) = 3 \\ \mathrm { f } ( 1 ) = 2 & \mathrm { f } ( 2 ) = 1 & \mathrm { f } ( 3 ) = 3 \end{array}$$ and the group operation is composition of functions.
  4. Show that ad \(= \mathrm { c }\) and find da.
  5. Give a reason why \(S\) is not isomorphic to \(G\).
  6. Find the order of each element of \(S\).
  7. List all the proper subgroups of \(S\).
Question 4:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
In \(G\): \(3^2=2,\ 3^3=6,\ 3^4=4,\ 3^5=5,\ 3^6=1\) [or \(5^2=4,\ 5^3=6,\ 5^4=2,\ 5^5=3,\ 5^6=1\)]M1 All powers of an element of order 6
In \(H\): \(5^2=7,\ 5^3=17,\ 5^4=13,\ 5^5=11,\ 5^6=1\) [or \(11^2=13,\ 11^3=17,\ 11^4=7,\ 11^5=5,\ 11^6=1\)]A1 All powers correct in both groups
\(G\) has an element 3 (or 5) of order 6B1
\(H\) has an element 5 (or 11) of order 6B1
Total: 4
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\{1,\ 6\}\)B1 Ignore \(\{1\}\) and \(G\)
\(\{1,\ 2,\ 4\}\)B2 Deduct 1 mark (from B1B2) for each proper subgroup in excess of two
Total: 3
Part (iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(G \leftrightarrow H\): \(1\leftrightarrow1,\ 2\leftrightarrow7,\ 3\leftrightarrow5,\ 4\leftrightarrow13,\ 5\leftrightarrow11,\ 6\leftrightarrow17\) OR \(1\leftrightarrow1,\ 2\leftrightarrow13,\ 3\leftrightarrow11,\ 4\leftrightarrow7,\ 5\leftrightarrow5,\ 6\leftrightarrow17\)B4 Give B3 for 4 correct, B2 for 3 correct, B1 for 2 correct
Total: 4
Part (iv):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(ad(1)=a(3)=1\), \(ad(2)=a(2)=3\), \(ad(3)=a(1)=2\), so \(ad=c\)M1, A1 Evaluating e.g. \(ad(1)\) (one case sufficient; intermediate value must be shown). For \(ad=c\) correctly shown
\(da(1)=d(2)=2\), \(da(2)=d(3)=1\), \(da(3)=d(1)=3\), so \(da=f\)M1, A1 Evaluating e.g. \(da(1)\) (one case sufficient; no need for any working)
Total: 4
Part (v):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(S\) is not abelian; \(G\) is abelianB1 or \(S\) has 3 elements of order 2; \(G\) has 1 element of order 2; or \(S\) is not cyclic, etc.
Total: 1
Part (vi):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Element: \(a,\ b,\ c,\ d,\ e,\ f\); Order: \(3,\ 3,\ 2,\ 2,\ 1,\ 2\)B4 Give B3 for 5 correct, B2 for 3 correct, B1 for 1 correct
Total: 4
Part (vii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\{e,c\},\ \{e,d\},\ \{e,f\}\)B1B1B1 Ignore \(\{e\}\) and \(S\)
\(\{e,a,b\}\)B1 If more than 4 proper subgroups given, deduct 1 mark for each in excess of 4
Total: 4 
# Question 4:

## Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| In $G$: $3^2=2,\ 3^3=6,\ 3^4=4,\ 3^5=5,\ 3^6=1$ [or $5^2=4,\ 5^3=6,\ 5^4=2,\ 5^5=3,\ 5^6=1$] | M1 | All powers of an element of order 6 |
| In $H$: $5^2=7,\ 5^3=17,\ 5^4=13,\ 5^5=11,\ 5^6=1$ [or $11^2=13,\ 11^3=17,\ 11^4=7,\ 11^5=5,\ 11^6=1$] | A1 | All powers correct in both groups |
| $G$ has an element 3 (or 5) of order 6 | B1 | |
| $H$ has an element 5 (or 11) of order 6 | B1 | |
| **Total: 4** | | |

## Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\{1,\ 6\}$ | B1 | Ignore $\{1\}$ and $G$ |
| $\{1,\ 2,\ 4\}$ | B2 | Deduct 1 mark (from B1B2) for each proper subgroup in excess of two |
| **Total: 3** | | |

## Part (iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $G \leftrightarrow H$: $1\leftrightarrow1,\ 2\leftrightarrow7,\ 3\leftrightarrow5,\ 4\leftrightarrow13,\ 5\leftrightarrow11,\ 6\leftrightarrow17$ OR $1\leftrightarrow1,\ 2\leftrightarrow13,\ 3\leftrightarrow11,\ 4\leftrightarrow7,\ 5\leftrightarrow5,\ 6\leftrightarrow17$ | B4 | Give B3 for 4 correct, B2 for 3 correct, B1 for 2 correct |
| **Total: 4** | | |

## Part (iv):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $ad(1)=a(3)=1$, $ad(2)=a(2)=3$, $ad(3)=a(1)=2$, so $ad=c$ | M1, A1 | Evaluating e.g. $ad(1)$ (one case sufficient; intermediate value must be shown). For $ad=c$ correctly shown |
| $da(1)=d(2)=2$, $da(2)=d(3)=1$, $da(3)=d(1)=3$, so $da=f$ | M1, A1 | Evaluating e.g. $da(1)$ (one case sufficient; no need for any working) |
| **Total: 4** | | |

## Part (v):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $S$ is not abelian; $G$ is abelian | B1 | or $S$ has 3 elements of order 2; $G$ has 1 element of order 2; or $S$ is not cyclic, etc. |
| **Total: 1** | | |

## Part (vi):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Element: $a,\ b,\ c,\ d,\ e,\ f$; Order: $3,\ 3,\ 2,\ 2,\ 1,\ 2$ | B4 | Give B3 for 5 correct, B2 for 3 correct, B1 for 1 correct |
| **Total: 4** | | |

## Part (vii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\{e,c\},\ \{e,d\},\ \{e,f\}$ | B1B1B1 | Ignore $\{e\}$ and $S$ |
| $\{e,a,b\}$ | B1 | If more than 4 proper subgroups given, deduct 1 mark for each in excess of 4 |
| **Total: 4** | | |

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4 The group $G = \{ 1,2,3,4,5,6 \}$ has multiplication modulo 7 as its operation. The group $H = \{ 1,5,7,11,13,17 \}$ has multiplication modulo 18 as its operation.\\
(i) Show that the groups $G$ and $H$ are both cyclic.\\
(ii) List all the proper subgroups of $G$.\\
(iii) Specify an isomorphism between $G$ and $H$.

The group $S = \{ \mathrm { a } , \mathrm { b } , \mathrm { c } , \mathrm { d } , \mathrm { e } , \mathrm { f } \}$ consists of functions with domain $\{ 1,2,3 \}$ given by

$$\begin{array} { l l l } 
\mathrm { a } ( 1 ) = 2 & \mathrm { a } ( 2 ) = 3 & \mathrm { a } ( 3 ) = 1 \\
\mathrm {~b} ( 1 ) = 3 & \mathrm {~b} ( 2 ) = 1 & \mathrm {~b} ( 3 ) = 2 \\
\mathrm { c } ( 1 ) = 1 & \mathrm { c } ( 2 ) = 3 & \mathrm { c } ( 3 ) = 2 \\
\mathrm {~d} ( 1 ) = 3 & \mathrm {~d} ( 2 ) = 2 & \mathrm {~d} ( 3 ) = 1 \\
\mathrm { e } ( 1 ) = 1 & \mathrm { e } ( 2 ) = 2 & \mathrm { e } ( 3 ) = 3 \\
\mathrm { f } ( 1 ) = 2 & \mathrm { f } ( 2 ) = 1 & \mathrm { f } ( 3 ) = 3
\end{array}$$

and the group operation is composition of functions.\\
(iv) Show that ad $= \mathrm { c }$ and find da.\\
(v) Give a reason why $S$ is not isomorphic to $G$.\\
(vi) Find the order of each element of $S$.\\
(vii) List all the proper subgroups of $S$.

\hfill \mbox{\textit{OCR MEI FP3 2009 Q4 [24]}}
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