OCR MEI Further Extra Pure 2024 June — Question 3 12 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
Year2024
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeGroups of symmetries
DifficultyChallenging +1.2 This is an introductory question on the dihedral group D₃, a standard Further Maths topic. Part (a) asks students to illustrate compositions using provided diagrams, which is primarily visual/mechanical work requiring understanding of the definitions but minimal problem-solving. While group theory is conceptually sophisticated, this specific question involves straightforward application of given transformations with visual aids, making it easier than average A-level questions overall but still requiring careful geometric reasoning.
Spec7.01d Multiplicative principle: arrangements of n distinct objects8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group

3 Fig. 3.1 shows an equilateral triangle, with vertices \(\mathrm { A } , \mathrm { B }\) and C , and the three axes of symmetry of the triangle, \(\mathrm { S } _ { \mathrm { a } } , \mathrm { S } _ { \mathrm { b } }\) and \(\mathrm { S } _ { \mathrm { c } }\). The axes of symmetry are fixed in space and all intersect at the point O . \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Fig. 3.1} \includegraphics[alt={},max width=\textwidth]{33c9e321-6044-45c4-bf37-0a6da3ecaf0d-4_440_394_440_248}
\end{figure} There are six distinct transformations under which the image of the triangle is indistinguishable from the triangle itself, ignoring labels.
These are denoted by \(\mathrm { I } , \mathrm { M } _ { a ^ { \prime } } \mathrm { M } _ { \mathrm { b } ^ { \prime } } , \mathrm { M } _ { \mathrm { c } ^ { \prime } } , \mathrm { R } _ { 120 }\) and \(\mathrm { R } _ { 240 }\) where
  • I is the identity transformation
  • \(\mathrm { M } _ { \mathrm { a } }\) is a reflection in the mirror line \(\mathrm { S } _ { \mathrm { a } }\) (and likewise for \(\mathrm { M } _ { \mathrm { b } }\) and \(\mathrm { M } _ { \mathrm { c } }\) )
  • \(\mathrm { R } _ { 120 }\) is an anticlockwise rotation by \(120 ^ { \circ }\) about O (and likewise for \(\mathrm { R } _ { 240 }\) ).
Composition of transformations is denoted by ○.
Fig. 3.2 illustrates the composition of \(R _ { 120 }\) followed by \(R _ { 240 }\), denoted by \(R _ { 240 } \circ R _ { 120 }\). This shows that \(\mathrm { R } _ { 240 } \circ \mathrm { R } _ { 120 }\) is equivalent to the identity transformation, so that \(\mathrm { R } _ { 240 } \circ \mathrm { R } _ { 120 } = \mathrm { I }\). \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Fig. 3.2} \includegraphics[alt={},max width=\textwidth]{33c9e321-6044-45c4-bf37-0a6da3ecaf0d-4_321_1447_1628_242}
\end{figure}
  1. Using the blank diagrams in the Printed Answer Booklet, find the single transformation which is equivalent to each of the following.
    The set of the six transformations is denoted by G and you are given that \(( \mathrm { G } , \circ )\) is a group. The table below is a mostly empty composition table for \(\circ\). The entry given is that for \(R _ { 240 } \circ R _ { 120 }\).
    First transformation performed is
    followed by
    I\(\mathrm { M } _ { \mathrm { a } }\)\(\mathrm { M } _ { \mathrm { b } }\)\(\mathrm { M } _ { \mathrm { c } }\)\(\mathrm { R } _ { 120 }\)\(\mathrm { R } _ { 240 }\)
    I
    \(\mathrm { M } _ { \mathrm { a } }\)
    \(\mathrm { M } _ { \mathrm { b } }\)
    \(\mathrm { M } _ { \mathrm { c } }\)
    \(\mathrm { R } _ { 120 }\)
    \(\mathrm { R } _ { 240 }\)I
  2. Complete the copy of this table in the Printed Answer Booklet. You can use some or all of the spare copies of the diagram in the Printed Answer Booklet to help.
  3. Explain why there can be no subgroup of \(( \mathrm { G } , \circ )\) of order 4.
  4. A student makes the following claim.
    "If all the proper non-trivial subgroups of a group are abelian then the group itself is abelian."
    Explain why the claim is incorrect, justifying your answer fully.
  5. With reference to the order of elements in the groups, explain why ( \(\mathrm { G } , \circ\) ) is not isomorphic to \(\mathrm { C } _ { 6 }\), the cyclic group of order 6 .

Question 3:
AnswerMarks Guidance
3(a) B1
B1
B1
AnswerMarks
[3]2.2a
2.2a
AnswerMarks
2.2aTriangles need not be labelled for
this B1 but must be correct if shown.
Triangles must be correctly labelled.
AnswerMarks
Triangles must be correctly labelled.The transformations do not need to be
shown on the arrows throughout.
If B?B0B0 then SCB1 if last two both
correct but triangles not labelled.
AnswerMarks Guidance
3(b)  I M a M b M c R 120 R 240
I I M M M R R
a b c 120 240
M M I R R M M
a a 120 240 b c
M M R I R M M
b b 240 120 c a
M M R R I M M
c c 120 240 a b
R R M M M R I
120 120 c a b 240
R R M M M I R
AnswerMarks
240 240 b c a 120B1
B1
B1
B1
AnswerMarks
[4]2.2a
2.2a
2.2a
AnswerMarks
2.2aIdentity row, column and leading
diagonal for reflections all correct.
Rotation block all correct.
Reflection combinations correct.
Rotation/reflection combinations
AnswerMarks
correct.If B0B0 for last two B marks then
SCB1 for fully correct table reflected
in leading diagonal.
AnswerMarks Guidance
3(c) The order of (G,) is 6. (By Lagrange’s theorem)
the order of any subgroup of (G,)must be a factor
AnswerMarks Guidance
of 6. 4 is not a factor of 6.B1
[1]2.2a or by Lagrange’s theorem the order
of any subgroup must be 1, 2, 3 (or 6)
and so the order of a subgroup can’t
be 4.
All three elements of the explanation
must be present.
Lagrange theorem must be stated or
appealed to, not inferred.
AnswerMarks
B0 if incorrect statement seen.Allow, “the order of G is 6, 4 is not a
factor of 6 so by Lagrange, 4 cannot be
the order of a subgroup.”
Condone, “The order of a subgroup
must be a factor of the order of the
group. 4 is not a factor of 6.”
AnswerMarks Guidance
I M
aM
bM
cR
120R
240
AnswerMarks Guidance
II M
aM
bM
cR
120R
240
M
AnswerMarks Guidance
aM
aI R
120R
240M
bM
c
M
AnswerMarks Guidance
bM
bR
240I R
120M
cM
a
M
AnswerMarks Guidance
cM
cR
120R
240I M
aM
b
R
AnswerMarks
120R
120M
cM
aM
bR
240I
R
AnswerMarks Guidance
240R
240M
bM
cM
aI R
120
AnswerMarks Guidance
3(d) From Lagrange’s theorem, all proper subgroups of
(G,) must have order 2 or 3 (or 1).
eg All of the proper non-trivial subgroups of (G,)
are abelian since they are all of order 2 or 3 and so
are abelian.
But (G,)is not abelian since the table is not
symmetric (about its leading diagonal)
AnswerMarks
and so the claim is incorrect.B1
B1
B1
AnswerMarks
[3]2.1
2.4
AnswerMarks
2.2aOr these can be explicitly listed (in
set or subgroup form) and stated as
being the only proper (non-trivial)
subgroups.
Ignore reference to ({I}, ).
Or explicit example, eg
R =M M ≠M M =R
240 a c c a 120
Indication is required of
understanding what it means for a
AnswerMarks
group to be non-abelian.{I, M }, {I, M }, {I, M } and
a b c
{I, R , R }. Condone e used for I.
120 240
The presence of {I} can be ignored
Could use < > notation.
`
Further guidance for marking:
B1 This group can only have (proper,
non-trivial) subgroups of order 2 or 3
(justified by Lagrange or complete list
explicitly given).
B1 Groups of order 2&3 are
necessarily abelian.
B1 This group is not abelian (justified
by example or table asymmetry).
AnswerMarks Guidance
3(e) (G,)has no elements of order 6. C has (some or
6
two or (at least) one) elements of order 6 (and
hence there can be no isomorphism between the
AnswerMarks Guidance
two groups.)B1
[1]2.2a Or (G,)has three elements of order
2 while C has one.
6
Answer must contain statements
relating to numerical order of
element(s) in both groups.
AnswerMarks
B0 if incorrect statement seen.NB Answer must use order of
elements, not order of subgroups or
other explanation (eg “not abelian =>
not cyclic”).
Not just “The orders of elements in the
group do not match” but OK if
supported by eg 1 2 3 6
G 1 3 2 0
C 1 1 2 2
6
Not just “No element generates G”.
Question 3:
3 | (a) | B1
B1
B1
[3] | 2.2a
2.2a
2.2a | Triangles need not be labelled for
this B1 but must be correct if shown.
Triangles must be correctly labelled.
Triangles must be correctly labelled. | The transformations do not need to be
shown on the arrows throughout.
If B?B0B0 then SCB1 if last two both
correct but triangles not labelled.
3 | (b) |  I M a M b M c R 120 R 240
I I M M M R R
a b c 120 240
M M I R R M M
a a 120 240 b c
M M R I R M M
b b 240 120 c a
M M R R I M M
c c 120 240 a b
R R M M M R I
120 120 c a b 240
R R M M M I R
240 240 b c a 120 | B1
B1
B1
B1
[4] | 2.2a
2.2a
2.2a
2.2a | Identity row, column and leading
diagonal for reflections all correct.
Rotation block all correct.
Reflection combinations correct.
Rotation/reflection combinations
correct. | If B0B0 for last two B marks then
SCB1 for fully correct table reflected
in leading diagonal.
3 | (c) | The order of (G,) is 6. (By Lagrange’s theorem)
the order of any subgroup of (G,)must be a factor
of 6. 4 is not a factor of 6. | B1
[1] | 2.2a | or by Lagrange’s theorem the order
of any subgroup must be 1, 2, 3 (or 6)
and so the order of a subgroup can’t
be 4.
All three elements of the explanation
must be present.
Lagrange theorem must be stated or
appealed to, not inferred.
B0 if incorrect statement seen. | Allow, “the order of G is 6, 4 is not a
factor of 6 so by Lagrange, 4 cannot be
the order of a subgroup.”
Condone, “The order of a subgroup
must be a factor of the order of the
group. 4 is not a factor of 6.”
 | I | M
a | M
b | M
c | R
120 | R
240
I | I | M
a | M
b | M
c | R
120 | R
240
M
a | M
a | I | R
120 | R
240 | M
b | M
c
M
b | M
b | R
240 | I | R
120 | M
c | M
a
M
c | M
c | R
120 | R
240 | I | M
a | M
b
R
120 | R
120 | M
c | M
a | M
b | R
240 | I
R
240 | R
240 | M
b | M
c | M
a | I | R
120
3 | (d) | From Lagrange’s theorem, all proper subgroups of
(G,) must have order 2 or 3 (or 1).
eg All of the proper non-trivial subgroups of (G,)
are abelian since they are all of order 2 or 3 and so
are abelian.
But (G,)is not abelian since the table is not
symmetric (about its leading diagonal)
and so the claim is incorrect. | B1
B1
B1
[3] | 2.1
2.4
2.2a | Or these can be explicitly listed (in
set or subgroup form) and stated as
being the only proper (non-trivial)
subgroups.
Ignore reference to ({I}, ).
Or explicit example, eg
R =M M ≠M M =R
240 a c c a 120
Indication is required of
understanding what it means for a
group to be non-abelian. | {I, M }, {I, M }, {I, M } and
a b c
{I, R , R }. Condone e used for I.
120 240
The presence of {I} can be ignored
Could use < > notation.
`
Further guidance for marking:
B1 This group can only have (proper,
non-trivial) subgroups of order 2 or 3
(justified by Lagrange or complete list
explicitly given).
B1 Groups of order 2&3 are
necessarily abelian.
B1 This group is not abelian (justified
by example or table asymmetry).
3 | (e) | (G,)has no elements of order 6. C has (some or
6
two or (at least) one) elements of order 6 (and
hence there can be no isomorphism between the
two groups.) | B1
[1] | 2.2a | Or (G,)has three elements of order
2 while C has one.
6
Answer must contain statements
relating to numerical order of
element(s) in both groups.
B0 if incorrect statement seen. | NB Answer must use order of
elements, not order of subgroups or
other explanation (eg “not abelian =>
not cyclic”).
Not just “The orders of elements in the
group do not match” but OK if
supported by eg 1 2 3 6
G 1 3 2 0
C 1 1 2 2
6
Not just “No element generates G”.
3 Fig. 3.1 shows an equilateral triangle, with vertices $\mathrm { A } , \mathrm { B }$ and C , and the three axes of symmetry of the triangle, $\mathrm { S } _ { \mathrm { a } } , \mathrm { S } _ { \mathrm { b } }$ and $\mathrm { S } _ { \mathrm { c } }$. The axes of symmetry are fixed in space and all intersect at the point O .

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
  \includegraphics[alt={},max width=\textwidth]{33c9e321-6044-45c4-bf37-0a6da3ecaf0d-4_440_394_440_248}
\end{center}
\end{figure}

There are six distinct transformations under which the image of the triangle is indistinguishable from the triangle itself, ignoring labels.\\
These are denoted by $\mathrm { I } , \mathrm { M } _ { a ^ { \prime } } \mathrm { M } _ { \mathrm { b } ^ { \prime } } , \mathrm { M } _ { \mathrm { c } ^ { \prime } } , \mathrm { R } _ { 120 }$ and $\mathrm { R } _ { 240 }$ where

\begin{itemize}
  \item I is the identity transformation
  \item $\mathrm { M } _ { \mathrm { a } }$ is a reflection in the mirror line $\mathrm { S } _ { \mathrm { a } }$ (and likewise for $\mathrm { M } _ { \mathrm { b } }$ and $\mathrm { M } _ { \mathrm { c } }$ )
  \item $\mathrm { R } _ { 120 }$ is an anticlockwise rotation by $120 ^ { \circ }$ about O (and likewise for $\mathrm { R } _ { 240 }$ ).
\end{itemize}

Composition of transformations is denoted by ○.\\
Fig. 3.2 illustrates the composition of $R _ { 120 }$ followed by $R _ { 240 }$, denoted by $R _ { 240 } \circ R _ { 120 }$. This shows that $\mathrm { R } _ { 240 } \circ \mathrm { R } _ { 120 }$ is equivalent to the identity transformation, so that $\mathrm { R } _ { 240 } \circ \mathrm { R } _ { 120 } = \mathrm { I }$.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
  \includegraphics[alt={},max width=\textwidth]{33c9e321-6044-45c4-bf37-0a6da3ecaf0d-4_321_1447_1628_242}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Using the blank diagrams in the Printed Answer Booklet, find the single transformation which is equivalent to each of the following.

\begin{itemize}
  \item $M _ { a } \circ M _ { a }$
  \item $M _ { b } \circ M _ { a }$
  \item $\mathrm { R } _ { 120 } \circ \mathrm { M } _ { \mathrm { a } }$
\end{itemize}

The set of the six transformations is denoted by G and you are given that $( \mathrm { G } , \circ )$ is a group.

The table below is a mostly empty composition table for $\circ$. The entry given is that for $R _ { 240 } \circ R _ { 120 }$.\\
First transformation performed is\\
followed by

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
◯ & I & $\mathrm { M } _ { \mathrm { a } }$ & $\mathrm { M } _ { \mathrm { b } }$ & $\mathrm { M } _ { \mathrm { c } }$ & $\mathrm { R } _ { 120 }$ & $\mathrm { R } _ { 240 }$ \\
\hline
I &  &  &  &  &  &  \\
\hline
$\mathrm { M } _ { \mathrm { a } }$ &  &  &  &  &  &  \\
\hline
$\mathrm { M } _ { \mathrm { b } }$ &  &  &  &  &  &  \\
\hline
$\mathrm { M } _ { \mathrm { c } }$ &  &  &  &  &  &  \\
\hline
$\mathrm { R } _ { 120 }$ &  &  &  &  &  &  \\
\hline
$\mathrm { R } _ { 240 }$ &  &  &  &  & I &  \\
\hline
\end{tabular}
\end{center}
\item Complete the copy of this table in the Printed Answer Booklet. You can use some or all of the spare copies of the diagram in the Printed Answer Booklet to help.
\item Explain why there can be no subgroup of $( \mathrm { G } , \circ )$ of order 4.
\item A student makes the following claim.\\
"If all the proper non-trivial subgroups of a group are abelian then the group itself is abelian."\\
Explain why the claim is incorrect, justifying your answer fully.
\item With reference to the order of elements in the groups, explain why ( $\mathrm { G } , \circ$ ) is not isomorphic to $\mathrm { C } _ { 6 }$, the cyclic group of order 6 .
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure 2024 Q3 [12]}}