A2 October 2020 Q6

EdexcelCurrent spec10 marksGroups

Figure 3: six-pointed star with vertices labelled 1 (top), 2, 3, 4 (bottom), 5, 6 anticlockwise; the triangles at vertices 1, 3 and 5 are shaded — Figure 3

Figure 3 shows a plane shape made up of a regular hexagon with an equilateral triangle joined to each edge and with alternate equilateral triangles shaded.

The symmetries of this shape are the rotations and reflections of the plane that preserve the shape and its shading.

The symmetries of the shape can be represented by permutations of the six vertices labelled 1 to 6 in Figure 3. The set of these permutations with the operation of composition form a group, \(G\).

(a) Describe geometrically the symmetry of the shape represented by the permutation\[\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 4 & 5 & 6 & 1 & 2 \end{pmatrix}\] (2)

(b) Write down, in similar two-line notation, the remaining elements of the group \(G\). (4)

(i) \(G\) has a subgroup of order 4

(ii) \(G\) is cyclic.

(2)

Diagram 1, below, shows an unshaded shape with the same outline as the shape in Figure 3.

Diagram 1: unshaded six-pointed star with the same outline as Figure 3 — Diagram 1

(d) Shade the shape in Diagram 1 in such a way that the group of symmetries of the resulting shaded shape is isomorphic to the cyclic group of order 6 (2)

Mark scheme (a)

Scheme	Marks	AO
A rotation \(\ldots\)	M1	1.1b
\(\ldots\) about the centre of the shape, though an angle 120° anticlockwise.	A1	1.1b
	(2)

Notes

M1: For identifying the permutation as representing a rotation.

A1: A complete description including rotation, centre, angle (degrees or radians) and direction.

Mark scheme (b)

Scheme	Marks	AO
\(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 6 & 1 & 2 & 3 & 4 \end{pmatrix}\)	B1	1.1b
One of \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 6 & 5 & 4 & 3 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 6 & 5 & 4 \end{pmatrix}\) or \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 4 & 3 & 2 & 1 & 6 \end{pmatrix}\)	M1	1.1b
Two of \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 6 & 5 & 4 & 3 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 6 & 5 & 4 \end{pmatrix}\) & \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 4 & 3 & 2 & 1 & 6 \end{pmatrix}\)	A1	1.1b
All the above and \(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \end{pmatrix}\) and no extra symmetries given. Two line form required for this mark.	A1	2.5
	(4)

Notes

NB accept in cycle form for the first two marks in (b).

B1: For giving the other rotation of order 3 (rotation through 240° anticlockwise).

M1: For one of the three reflections correctly given.

A1: For one of the other two reflections.

A1: For all reflections and the identity, and no extra symmetries given, and all in two-line notation.

Mark scheme (c)

Scheme	Marks	AO
\(G\) has order 6 so can have no subgroup of order 4 by Lagrange’s Theorem.	B1	2.4
There is no element of order 6 that generates the group.	B1	2.4
	(2)

Notes

B1: Correct reason given. Stating there are no elements of order 4 is not sufficient. A longer, valid method would be to state a subgroup of order 4 would need to comprise of the identity and the three elements of order 2 (as it cannot contain any of order 3), but this is not closed as the composite of any two reflection is a rotation.

B1: Refers to there being no element (of order 6) to generate the group. May reason using orders of elements or via geometric restrictions (e.g. cannot generate a rotation from a single reflection and vice versa).