Question 5 - A-Level Maths

OCR Further Additional Pure 2019 June — Question 5 11 marks

Exam Board	OCR
Module	Further Additional Pure (Further Additional Pure)
Year	2019
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Groups
Type	Isomorphism between groups
Difficulty	Challenging +1.2 This is a structured group theory question requiring construction of a cyclic group generated by 11 mod 80, comparison of Cayley tables, and identification of subgroups. While it involves multiple parts and abstract algebra concepts beyond standard A-level, the tasks are methodical and guided: computing powers of 11 mod 80 is computational, comparing group structures via Cayley tables is direct observation, and finding subgroups follows from identifying elements' orders. The question requires familiarity with group theory terminology but doesn't demand deep insight or novel problem-solving approaches—it's a standard Further Maths group theory exercise.
Spec	8.03b Cayley tables: construct for finite sets under binary operation 8.03c Group definition: recall and use, show structure is/isn't a group 8.03g Cyclic groups: meaning of the term 8.03h Generators: of cyclic and non-cyclic groups 8.03l Isomorphism: determine using informal methods

5 The group \(G\) consists of a set \(S\) together with \(\times _ { 80 }\), the operation of multiplication modulo 80. It is given that \(S\) is the smallest set which contains the element 11 .

By constructing the Cayley table for \(G\), determine all the elements of \(S\). The Cayley table for a second group, \(H\), also with the operation \(\times _ { 80 }\), is shown below.
\cline { 2 - 5 } \multicolumn{1}{c|}{\(\times _ { 80 }\)} 1 9 31 39
1 1 9 31 39
9 9 1 39 31
31 31 39 1 9
39 39 31 9 1
Use the two Cayley tables to explain why \(G\) and \(H\) are not isomorphic.
(i) List

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
5	(a)	112 =121 ≡ 41 (mod 80) so 41 ∈ S

11 × 41 = 451 ≡ 51 (mod 80) so 51 ∈ S

412 ≡ 1 (mod 80) so 1 ∈ S

× 1 11 41 51

80 1 1 11 41 51

11 11 41 51 1

41 41 51 1 11

Answer	Marks
51 51 1 11 41	3.1a

2.5 2.2a

1.1a

Answer	Marks
1.1	B1

B1

M1

A1

Answer	Marks
[5]	B1 for each other element found (possibly tabulated)

4×4 table attempted (with at least R and C correct) using the

1 1

correct 4 elements

All correct

Answer	Marks
(b)	Choice of a suitable method, such as

Listing orders of elements: 1, 4, 2, 4 in G and 1, 2, 2, 2 in H

OR Stating that G is the cyclic group of order 4 while H is the Klein-4 group

OR Stating that G has an element of order 4 while H has all (non-identity)

Answer	Marks
elements of order 2	2.1
2.4	M1

A1

Answer	Marks
[2]	MUST be for two groups of order 4

Using knowledge of small-order groups: M1 for clear indication

that there are only two groups of order 4, A1 for the details

Answer	Marks	Guidance
(c)	i	G has proper subgroup {1, 41} only
H has proper subgroups {1, 9}, {1, 31} and {1, 39}	3.1a

1.2

Answer	Marks
2.2a	B1

M1

A1

Answer	Marks
[3]	Ignore statement of {1} and group throughout

Listing any two correct subgroups of order 2

All three and no extras

Answer	Marks	Guidance
ii	G and H must have different structures since they have differing numbers of
(proper) subgroups	2.4	B1
[1]	(Must have different numbers of subgroups in (c)(i))
1	11	41

Question 5:
5 | (a) | 112 =121 ≡ 41 (mod 80) so 41 ∈ S
11 × 41 = 451 ≡ 51 (mod 80) so 51 ∈ S
412 ≡ 1 (mod 80) so 1 ∈ S
× 1 11 41 51
80
1 1 11 41 51
11 11 41 51 1
41 41 51 1 11
51 51 1 11 41 | 3.1a
2.5
2.2a
1.1a
1.1 | B1
B1
B1
M1
A1
[5] | B1 for each other element found (possibly tabulated)
4×4 table attempted (with at least R and C correct) using the
1 1
correct 4 elements
All correct
(b) | Choice of a suitable method, such as
Listing orders of elements: 1, 4, 2, 4 in G and 1, 2, 2, 2 in H
OR Stating that G is the cyclic group of order 4 while H is the Klein-4 group
OR Stating that G has an element of order 4 while H has all (non-identity)
elements of order 2 | 2.1
2.4 | M1
A1
[2] | MUST be for two groups of order 4
Using knowledge of small-order groups: M1 for clear indication
that there are only two groups of order 4, A1 for the details
(c) | i | G has proper subgroup {1, 41} only
H has proper subgroups {1, 9}, {1, 31} and {1, 39} | 3.1a
1.2
2.2a | B1
M1
A1
[3] | Ignore statement of {1} and group throughout
Listing any two correct subgroups of order 2
All three and no extras
ii | G and H must have different structures since they have differing numbers of
(proper) subgroups | 2.4 | B1
[1] | (Must have different numbers of subgroups in (c)(i))
1 | 11 | 41 | 51

Show LaTeX source

5 The group $G$ consists of a set $S$ together with $\times _ { 80 }$, the operation of multiplication modulo 80. It is given that $S$ is the smallest set which contains the element 11 .
\begin{enumerate}[label=(\alph*)]
\item By constructing the Cayley table for $G$, determine all the elements of $S$.

The Cayley table for a second group, $H$, also with the operation $\times _ { 80 }$, is shown below.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\cline { 2 - 5 }
\multicolumn{1}{c|}{$\times _ { 80 }$} & 1 & 9 & 31 & 39 \\
\hline
1 & 1 & 9 & 31 & 39 \\
\hline
9 & 9 & 1 & 39 & 31 \\
\hline
31 & 31 & 39 & 1 & 9 \\
\hline
39 & 39 & 31 & 9 & 1 \\
\hline
\end{tabular}
\end{center}
\item Use the two Cayley tables to explain why $G$ and $H$ are not isomorphic.
\item (i) List

\begin{itemize}
  \item all the proper subgroups of $G$,
  \item all the proper subgroups of $H$.\\
(ii) Use your answers to (c) (i) to give another reason why $G$ and $H$ are not isomorphic.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2019 Q5 [11]}}

This paper (8 questions)

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Q1 4 Q2 11 Q3 4 Q4 10 Q5 11 Q6 12 Q7 12 Q8 11

\cline { 2 - 5 } \multicolumn{1}{c\|}{\(\times _ { 80 }\)}	1	9	31	39
1	1	9	31	39
9	9	1	39	31
31	31	39	1	9
39	39	31	9	1