OCR MEI FP3 2012 June — Question 4 24 marks

Exam BoardOCR MEI
ModuleFP3 (Further Pure Mathematics 3)
Year2012
SessionJune
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeIsomorphism between groups
DifficultyChallenging +1.3 This is a structured group theory question with multiple parts building systematically. Parts (i)-(v) involve routine verification of group properties, inverse relationships, and constructing a composition table to show isomorphism—all standard FP3 techniques. Part (vi) requires reading orders from a given table. Part (vii) identifying subgroups of D₄ is more demanding but follows from systematic checking. While this covers substantial ground, each individual step is methodical rather than requiring novel insight, making it moderately above average difficulty for Further Maths.
Spec8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03l Isomorphism: determine using informal methods
4
  1. Show that the set \(P = \{ 1,5,7,11 \}\), under the binary operation of multiplication modulo 12, is a group. You may assume associativity. A group \(Q\) has identity element \(e\). The result of applying the binary operation of \(Q\) to elements \(x\) and \(y\) is written \(x y\), and the inverse of \(x\) is written \(x ^ { - 1 }\).
  2. Verify that the inverse of \(x y\) is \(y ^ { - 1 } x ^ { - 1 }\). Three elements \(a , b\) and \(c\) of \(Q\) all have order 2, and \(a b = c\).
  3. By considering the inverse of \(c\), or otherwise, show that \(b a = c\).
  4. Show that \(b c = a\) and \(a c = b\). Find \(c b\) and \(c a\).
  5. Complete the composition table for \(R = \{ e , a , b , c \}\). Hence show that \(R\) is a subgroup of \(Q\) and that \(R\) is isomorphic to \(P\). The group \(T\) of symmetries of a square contains four reflections \(A , B , C , D\), the identity transformation \(E\) and three rotations \(F , G , H\). The binary operation is composition of transformations. The composition table for \(T\) is given below.
    A\(B\)\(C\)D\(E\)\(F\)\(G\)\(H\)
    AE\(G\)\(H\)\(F\)\(A\)D\(B\)\(C\)
    BGE\(F\)\(H\)\(B\)CAD
    C\(F\)HEGCAD\(B\)
    D\(H\)\(F\)\(G\)E\(D\)\(B\)C\(A\)
    EA\(B\)CD\(E\)\(F\)\(G\)\(H\)
    FCD\(B\)A\(F\)G\(H\)\(E\)
    \(G\)B\(A\)\(D\)C\(G\)HE\(F\)
    \(H\)DCAB\(H\)E\(F\)G
  6. Find the order of each element of \(T\).
  7. List all the proper subgroups of \(T\).
Question 4:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
Cayley table for \(P = \{1, 5, 7, 11\}\) completed correctlyB1
Table shows closureB1
Identity is 1B1
All elements are self-inverseB1 Condone no mention of inverse of 1
[4]
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
\((xy)(y^{-1}x^{-1})\)M1 Or \((y^{-1}x^{-1})(xy)\)
\(= x(yy^{-1})x^{-1} = xex^{-1} = xx^{-1} = e\)E1
So \(y^{-1}x^{-1}\) is the inverse of \(xy\)
[2]
Part (iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(a^{-1}=a,\ b^{-1}=b,\ c^{-1}=c,\ c^{-1}=(ab)^{-1}=b^{-1}a^{-1}\)M1 For any one of these
Hence \(c = ba\)E1
[2]
Part (iv):
AnswerMarks Guidance
AnswerMarks Guidance
\(bc = b(ba)\)M1 Or \(ba=c \Rightarrow a=b^{-1}c\); any correct first step
\(bc = ea = a\)E1
\(ac = a(ab) = eb = b\)E1
\(cb = a,\ ca = b\)B1
[4]
Part (v):
AnswerMarks Guidance
AnswerMarks Guidance
Cayley table for \(R = \{e, a, b, c\}\) completed correctlyB1
\(R\) is closedM1
Hence \(R\) is a subgroupE1 No need to mention identity or inverses; dependent on B1 only
Same pattern as \(P\); hence \(R\) and \(P\) are isomorphicE1
[4]
Part (vi):
AnswerMarks Guidance
AnswerMarks Guidance
Orders: \(A\):2, \(B\):2, \(C\):2, \(D\):2, \(E\):1, \(F\):4, \(G\):2, \(H\):4B3 Give B1 for 3 correct; B2 for 6 correct
[3]
Part (vii):
AnswerMarks Guidance
AnswerMarks Guidance
Subgroups of order 2: \(\{E,A\},\{E,B\},\{E,C\},\{E,D\},\{E,G\}\)B2 Ignore \(\{E\}\) and \(T\); Give B1 for 3 correct; deduct 1 mark (from B2) for each subgroup of order 2 given in excess of five
\(\{E,F,G,H\}\)B1 Deduct 1 mark (from B1B1B1) for each subgroup of order 3 or more given in excess of three
\(\{E,A,B,G\}\)B1
\(\{E,C,D,G\}\)B1
[5] 
# Question 4:

## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley table for $P = \{1, 5, 7, 11\}$ completed correctly | B1 | |
| Table shows closure | B1 | |
| Identity is 1 | B1 | |
| All elements are self-inverse | B1 | Condone no mention of inverse of 1 |
| | **[4]** | |

## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(xy)(y^{-1}x^{-1})$ | M1 | Or $(y^{-1}x^{-1})(xy)$ |
| $= x(yy^{-1})x^{-1} = xex^{-1} = xx^{-1} = e$ | E1 | |
| So $y^{-1}x^{-1}$ is the inverse of $xy$ | | |
| | **[2]** | |

## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a^{-1}=a,\ b^{-1}=b,\ c^{-1}=c,\ c^{-1}=(ab)^{-1}=b^{-1}a^{-1}$ | M1 | For any one of these |
| Hence $c = ba$ | E1 | |
| | **[2]** | |

## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $bc = b(ba)$ | M1 | Or $ba=c \Rightarrow a=b^{-1}c$; any correct first step |
| $bc = ea = a$ | E1 | |
| $ac = a(ab) = eb = b$ | E1 | |
| $cb = a,\ ca = b$ | B1 | |
| | **[4]** | |

## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley table for $R = \{e, a, b, c\}$ completed correctly | B1 | |
| $R$ is closed | M1 | |
| Hence $R$ is a subgroup | E1 | No need to mention identity or inverses; dependent on B1 only |
| Same pattern as $P$; hence $R$ and $P$ are isomorphic | E1 | |
| | **[4]** | |

## Part (vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Orders: $A$:2, $B$:2, $C$:2, $D$:2, $E$:1, $F$:4, $G$:2, $H$:4 | B3 | Give B1 for 3 correct; B2 for 6 correct |
| | **[3]** | |

## Part (vii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Subgroups of order 2: $\{E,A\},\{E,B\},\{E,C\},\{E,D\},\{E,G\}$ | B2 | Ignore $\{E\}$ and $T$; Give B1 for 3 correct; deduct 1 mark (from B2) for each subgroup of order 2 given in excess of five |
| $\{E,F,G,H\}$ | B1 | Deduct 1 mark (from B1B1B1) for each subgroup of order 3 or more given in excess of three |
| $\{E,A,B,G\}$ | B1 | |
| $\{E,C,D,G\}$ | B1 | |
| | **[5]** | |

---
4 (i) Show that the set $P = \{ 1,5,7,11 \}$, under the binary operation of multiplication modulo 12, is a group. You may assume associativity.

A group $Q$ has identity element $e$. The result of applying the binary operation of $Q$ to elements $x$ and $y$ is written $x y$, and the inverse of $x$ is written $x ^ { - 1 }$.\\
(ii) Verify that the inverse of $x y$ is $y ^ { - 1 } x ^ { - 1 }$.

Three elements $a , b$ and $c$ of $Q$ all have order 2, and $a b = c$.\\
(iii) By considering the inverse of $c$, or otherwise, show that $b a = c$.\\
(iv) Show that $b c = a$ and $a c = b$. Find $c b$ and $c a$.\\
(v) Complete the composition table for $R = \{ e , a , b , c \}$. Hence show that $R$ is a subgroup of $Q$ and that $R$ is isomorphic to $P$.

The group $T$ of symmetries of a square contains four reflections $A , B , C , D$, the identity transformation $E$ and three rotations $F , G , H$. The binary operation is composition of transformations. The composition table for $T$ is given below.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
 & A & $B$ & $C$ & D & $E$ & $F$ & $G$ & $H$ \\
\hline
A & E & $G$ & $H$ & $F$ & $A$ & D & $B$ & $C$ \\
\hline
B & G & E & $F$ & $H$ & $B$ & C & A & D \\
\hline
C & $F$ & H & E & G & C & A & D & $B$ \\
\hline
D & $H$ & $F$ & $G$ & E & $D$ & $B$ & C & $A$ \\
\hline
E & A & $B$ & C & D & $E$ & $F$ & $G$ & $H$ \\
\hline
F & C & D & $B$ & A & $F$ & G & $H$ & $E$ \\
\hline
$G$ & B & $A$ & $D$ & C & $G$ & H & E & $F$ \\
\hline
$H$ & D & C & A & B & $H$ & E & $F$ & G \\
\hline
\end{tabular}
\end{center}

(vi) Find the order of each element of $T$.\\
(vii) List all the proper subgroups of $T$.

\hfill \mbox{\textit{OCR MEI FP3 2012 Q4 [24]}}
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