OCR MEI FP3 2016 June — Question 4 24 marks

Exam BoardOCR MEI
ModuleFP3 (Further Pure Mathematics 3)
Year2016
SessionJune
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeIsomorphism between groups
DifficultyChallenging +1.2 This is a comprehensive group theory question covering standard topics (closure/associativity verification, orders, subgroups, cyclic groups, composition tables, isomorphisms). While it requires multiple techniques and understanding of group axioms, each part follows textbook procedures without requiring novel insight. The isomorphism identification is straightforward (comparing cyclic vs Klein-4 structure), and part (d) is a simple counterexample proof. Slightly above average due to length and breadth, but all components are routine 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 group8.03d Latin square property: for group tables8.03e Order of elements: and order of groups8.03g Cyclic groups: meaning of the term8.03h Generators: of cyclic and non-cyclic groups8.03l Isomorphism: determine using informal methods

4
  1. The elements of the set \(P = \{ 1,3,9,11 \}\) are combined under the binary operation, *, defined as multiplication modulo 16.
    1. Demonstrate associativity for the elements \(3,9,11\) in that order. Assuming associativity holds in general, show that \(P\) forms a group under the binary operation *.
    2. Write down the order of each element.
    3. Write down all subgroups of \(P\).
    4. Show that the group in part (i) is cyclic.
  2. Now consider a group of order 4 containing the identity element \(e\) and the two distinct elements, \(a\) and \(b\), where \(a ^ { 2 } = b ^ { 2 } = e\). Construct the composition table. Show that the group is non-cyclic.
  3. Now consider the four matrices \(\mathbf { I } , \mathbf { X } , \mathbf { Y }\) and \(\mathbf { Z }\) where $$\mathbf { I } = \left( \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right) , \mathbf { X } = \left( \begin{array} { r r } 1 & 0 \\ 0 & - 1 \end{array} \right) , \mathbf { Y } = \left( \begin{array} { r r } - 1 & 0 \\ 0 & 1 \end{array} \right) , \mathbf { Z } = \left( \begin{array} { r r } - 1 & 0 \\ 0 & - 1 \end{array} \right) .$$ The group G consists of the set \(\{ \mathbf { I } , \mathbf { X } , \mathbf { Y } , \mathbf { Z } \}\) with binary operation matrix multiplication. Determine which of the groups in parts (a) and (b) is isomorphic to G, and specify the isomorphism.
  4. The distinct elements \(\{ p , q , r , s \}\) are combined under the binary operation \({ } ^ { \circ }\). You are given that \(p ^ { \circ } q = r\) and \(q ^ { \circ } p = s\). By reference to the group axioms, prove that \(\{ p , q , r , s \}\) is not a group under \({ } ^ { \circ }\). Option 5: Markov chains \section*{This question requires the use of a calculator with the ability to handle matrices.}

Question 4 (a)(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(3*(9*11) = 3*3 = 9\)B1 Group table shown
\((3*9)*11 = 11*11 = 9\)B1
Construction of group table (or otherwise)B1
It shows closureB1
The identity is 1B1
Each element has an inverse: \(3^{-1}=11\), \(9^{-1}=9\), \(11^{-1}=3\), \(1^{-1}=1\)B1
Question 4 (a)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
Element: \(1,\ 3,\ 9,\ 11\); Order: \(1,\ 4,\ 2,\ 4\)B2 \(-1\) each error
Question 4 (a)(iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\{1\}\), \(\{1,9\}\), \(\{1,3,9,11\}\)B1 Condone omission of trivial subgroups; B0 if any extras
Question 4 (a)(iv):
AnswerMarks Guidance
AnswerMarks Guidance
e.g. \(3^2=9\), \(3^3=11\), \(3^4=1\); 3 generates the group and so it is cyclicE1
Question 4 (b):
AnswerMarks Guidance
AnswerMarks Guidance
Composition table with elements \(e, a, b, ab\) correctly completedB3 \(-1\) each error
All elements are self-inverse, and so no element generates the groupE1
Question 4 (c):
AnswerMarks Guidance
AnswerMarks Guidance
In group G all elements are self-inverse, i.e. \(X^2=I\), \(Y^2=I\), \(Z^2=I\)M1, A1A1
So this group is isomorphic to the group in (b)A1 Correctly shown
e.g. \(I\leftrightarrow e\), \(X\leftrightarrow a\), \(Y\leftrightarrow b\), \(Z\leftrightarrow ab\)B1B1
Question 4 (d):
AnswerMarks Guidance
AnswerMarks Guidance
One of the elements needs to be the identity elementM1
It is neither p nor q, for otherwise \(p \circ q = p\) (or q)A1
It is neither r nor s, for otherwise \(p \circ q = q \circ p = r\) (or s)A1
So there is no identity element and so not a groupE1
# Question 4 (a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $3*(9*11) = 3*3 = 9$ | B1 | Group table shown |
| $(3*9)*11 = 11*11 = 9$ | B1 | |
| Construction of group table (or otherwise) | B1 | |
| It shows closure | B1 | |
| The identity is 1 | B1 | |
| Each element has an inverse: $3^{-1}=11$, $9^{-1}=9$, $11^{-1}=3$, $1^{-1}=1$ | B1 | |

---

# Question 4 (a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Element: $1,\ 3,\ 9,\ 11$; Order: $1,\ 4,\ 2,\ 4$ | B2 | $-1$ each error |

---

# Question 4 (a)(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\{1\}$, $\{1,9\}$, $\{1,3,9,11\}$ | B1 | Condone omission of trivial subgroups; B0 if any extras |

---

# Question 4 (a)(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. $3^2=9$, $3^3=11$, $3^4=1$; 3 generates the group and so it is cyclic | E1 | |

---

# Question 4 (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Composition table with elements $e, a, b, ab$ correctly completed | B3 | $-1$ each error |
| All elements are self-inverse, and so no element generates the group | E1 | |

---

# Question 4 (c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| In group G all elements are self-inverse, i.e. $X^2=I$, $Y^2=I$, $Z^2=I$ | M1, A1A1 | |
| So this group is isomorphic to the group in (b) | A1 | Correctly shown |
| e.g. $I\leftrightarrow e$, $X\leftrightarrow a$, $Y\leftrightarrow b$, $Z\leftrightarrow ab$ | B1B1 | |

---

# Question 4 (d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| One of the elements needs to be the identity element | M1 | |
| It is neither p nor q, for otherwise $p \circ q = p$ (or q) | A1 | |
| It is neither r nor s, for otherwise $p \circ q = q \circ p = r$ (or s) | A1 | |
| So there is no identity element and so not a group | E1 | |
4
\begin{enumerate}[label=(\alph*)]
\item The elements of the set $P = \{ 1,3,9,11 \}$ are combined under the binary operation, *, defined as multiplication modulo 16.
\begin{enumerate}[label=(\roman*)]
\item Demonstrate associativity for the elements $3,9,11$ in that order.

Assuming associativity holds in general, show that $P$ forms a group under the binary operation *.
\item Write down the order of each element.
\item Write down all subgroups of $P$.
\item Show that the group in part (i) is cyclic.
\end{enumerate}\item Now consider a group of order 4 containing the identity element $e$ and the two distinct elements, $a$ and $b$, where $a ^ { 2 } = b ^ { 2 } = e$. Construct the composition table. Show that the group is non-cyclic.
\item Now consider the four matrices $\mathbf { I } , \mathbf { X } , \mathbf { Y }$ and $\mathbf { Z }$ where

$$\mathbf { I } = \left( \begin{array} { l l } 
1 & 0 \\
0 & 1
\end{array} \right) , \mathbf { X } = \left( \begin{array} { r r } 
1 & 0 \\
0 & - 1
\end{array} \right) , \mathbf { Y } = \left( \begin{array} { r r } 
- 1 & 0 \\
0 & 1
\end{array} \right) , \mathbf { Z } = \left( \begin{array} { r r } 
- 1 & 0 \\
0 & - 1
\end{array} \right) .$$

The group G consists of the set $\{ \mathbf { I } , \mathbf { X } , \mathbf { Y } , \mathbf { Z } \}$ with binary operation matrix multiplication. Determine which of the groups in parts (a) and (b) is isomorphic to G, and specify the isomorphism.
\item The distinct elements $\{ p , q , r , s \}$ are combined under the binary operation ${ } ^ { \circ }$. You are given that $p ^ { \circ } q = r$ and $q ^ { \circ } p = s$.

By reference to the group axioms, prove that $\{ p , q , r , s \}$ is not a group under ${ } ^ { \circ }$.

Option 5: Markov chains

\section*{This question requires the use of a calculator with the ability to handle matrices.}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI FP3 2016 Q4 [24]}}