A2 June 2024 Q7

EdexcelCurrent spec10 marksGroups

7. The set of matrices \(G = \{\mathbf{I}, \mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D}, \mathbf{E}\}\) where

\[\mathbf{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad \mathbf{A} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad \mathbf{B} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \quad \mathbf{C} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \quad \mathbf{D} = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \quad \mathbf{E} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}\]

with the operation \(\otimes_2\) of matrix multiplication with entries evaluated modulo 2, forms a group.

(a) Show that \(\mathbf{B}\) is an element of order 3 in \(G\). (2)

(b) Determine the orders of the other elements of \(G\). (3)

(i) a cyclic group of order 6

(ii) the group of symmetries of a regular hexagon. (2)

The group \(H\) of permutations of the numbers 1, 2 and 3 contains the following elements, denoted in two-line notation,

\[e = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} \qquad a = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \qquad b = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}\]\[c = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} \qquad d = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix} \qquad f = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix}\]

(d) Determine an isomorphism between the groups \(G\) and \(H\). (3)

Mark scheme (a)

Scheme	Marks	AO
\(\mathbf{B}^2 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}\{= \mathbf{E}\}\) Accept \(\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\) as long as \(2 \equiv 0 \pmod{2}\) is used later. Or \(\mathbf{B}^3 = \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\{= \mathbf{I}\}\) Accept \(\begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix}\) as long as \(2 \equiv 0 \pmod{2}\) and \(3 \equiv 1 \pmod{2}\) is used later.	M1	2.5
\(\mathbf{B}^3 = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\{= \mathbf{I}\}\) (and B is not the identity) hence B has order 3 or \(\|\mathbf{B}\| = 3\)	A1	2.1
	(2)

Notes

M1: Squares B to begin the process of finding the order, or uses a calculator to find \(\mathbf{B}^3\)

A1: Shows that \(\mathbf{B}^3 = \mathbf{I}\) and concludes that order must be 3.

Mark scheme (b)

Scheme	Marks	AO
I {is the identity so} has order 1, and E {is \(\mathbf{B}^{-1}\) so also} has order 3.	B1	1.1b
Finds at least one of the remaining orders e.g. \(\mathbf{A}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\{= \mathbf{I}\}\) or \(\mathbf{C}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\{= \mathbf{I}\}\) or \(\mathbf{D}^2 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\{= \mathbf{I}\}\)	M1	1.1b
(So I has order 1, B and E order 3 and ) A, C and D each have order 2	A1	1.1b
	(3)

Notes

B1: States the order for I and gives the order for E (which is the inverse of B).

M1: Finds the order of at least one of the other elements to get an element of order 2. May use a calculator

A1: Correct orders for the three elements of order 2.

(corrected from the printed mark scheme: the printed working shows \(\mathbf{A}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf{D}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\), but \(\mathbf{A}^2\) is \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) directly and \(\mathbf{D}^2 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}\) before reducing modulo 2)

Mark scheme (c)

Scheme	Marks	AO
(i) There is no element of order 6 in \(G\) {so cannot be isomorphic to \(C_6\)}	B1	2.4
(ii) Either There are 12 symmetries of a regular hexagon Hexagon has more than 6 symmetries Hexagon has 2 rotations of order 6, could give an example Hexagon has 2 rotations of order 3 Hexagon has 1 rotation of order 2 Hexagon has 6 reflections of order 2 Group of symmetries of a hexagon has an order greater than 6 B0 for The group of symmetries of a hexagon is also cyclic The groups are not of the same order Any incorrect orders of element stated Comments on the order of the whole group The group of symmetries has order 6	B1	2.4
	(2)

Scheme

Marks

(i) There is no element of order 6 in \(G\) {so cannot be isomorphic to \(C_6\)}

2.4

(ii) Either

There are 12 symmetries of a regular hexagon
Hexagon has more than 6 symmetries
Hexagon has 2 rotations of order 6, could give an example
Hexagon has 2 rotations of order 3
Hexagon has 1 rotation of order 2
Hexagon has 6 reflections of order 2
Group of symmetries of a hexagon has an order greater than 6

B0 for

The group of symmetries of a hexagon is also cyclic
The groups are not of the same order
Any incorrect orders of element stated
Comments on the order of the whole group
The group of symmetries has order 6

2.4

(2)

Notes

B1: Correct reason given, identifies there is no element of order 6. It is not sufficient to say that it does not have a group generator, they need to give a reason.

B1: Any correct reason given.

Considers the total number of elements of the group, order of the group is more than 6

Considers the orders of either reflections or rotations

Mark scheme (d)

Scheme

Marks

Any two correct matchings

2.2a

Completes the matching with at least 4 correct

3.1a

Hence an isomorphism is

I	A	B	C	D	E
\(e\)	\(c\)	\(a\)	\(d\)	\(f\)	\(b\)
\(e\)	\(d\)	\(a\)	\(f\)	\(c\)	\(b\)
\(e\)	\(f\)	\(a\)	\(c\)	\(d\)	\(b\)

I	A	B	C	D	E
\(e\)	\(c\)	\(b\)	\(f\)	\(d\)	\(a\)
\(e\)	\(d\)	\(b\)	\(c\)	\(f\)	\(a\)
\(e\)	\(f\)	\(b\)	\(d\)	\(c\)	\(a\)

2.1

(3)

(10 marks)

Notes

M1: Any two correct matchings

M1: Completes the matching with at least 4 correct

A1: A fully specified correct isomorphism.