Groups

Topic

AS June 2025 Q2

EdexcelCurrent spec8 marksGroups Number Theory

(i) Using a suitable algorithm and without performing any division, determine whether 13 306 617 is divisible by 9 (2)

(ii) The group \(G = \{1, 3, 7, 9, 11, 13, 17, 19\}\) has multiplication modulo 20 as its operation.

(a) Complete the following Cayley table for \(G\)

\(\times_{20}\)	1	3	7	9	11	13	17	19
1	1	3	7	9	11	13	17	19
3	3		1			19	11
7	7	1			17			13
9	9		3	1		17	13
11	11			19	1
13	13		11			9	1	7
17	17	11						3
19	19		13		9		3

(3)

(b) State the inverse of the element 7 (1)

(d) Write down a subgroup of \(G\) of order 4 (1)

Mark scheme (i)

Scheme	Marks	AO
\(1 + 3 + 3 + 6 + 6 + 1 + 7 = 27\)	M1	1.1b
The sum of the digits is 27 which is a multiple of 9 so 13 306 617 is divisible by 9	A1	1.1b
	(2)

Notes

M1: Applies the divisibility test by summing the digits.

A1: Correct solution and explanation.

Scheme	Marks	AO
3	B1	1.1b
	(1)

Scheme	Marks	AO
4	B1	1.1b
	(1)

Scheme	Marks	AO
\(\{1, 3, 7, 9\}\) or \(\{1, 9, 13, 17\}\) or \(\{1, 9, 11, 19\}\)	B1	1.1b
	(1)
	(8 marks)

A2 June 2025 Q1

EdexcelCurrent spec7 marksGroups

1. The set \(S = \{1, 3, 5, 9, 11, 13\}\) forms the group \(G\), under the operation multiplication modulo 14

(a) Complete the Cayley table below for the group \(G\)

\(\times_{14}\)	1	3	5	9	11	13
1	1	3	5	9	11	13
3	3	9	1	13	5	11
5	5	1	11
9	9	13		11
11	11	5			9
13	13	11				1

(3)

(b) Write down a subgroup of \(G\) of order 2 (1)

The group \(H\) is defined by the Cayley table below.

\(*\)	\(p\)	\(q\)	\(r\)	\(s\)	\(t\)	\(u\)
\(p\)	\(p\)	\(q\)	\(r\)	\(s\)	\(t\)	\(u\)
\(q\)	\(q\)	\(t\)	\(u\)	\(r\)	\(s\)	\(p\)
\(r\)	\(r\)	\(u\)	\(t\)	\(q\)	\(p\)	\(s\)
\(s\)	\(s\)	\(r\)	\(q\)	\(p\)	\(u\)	\(t\)
\(t\)	\(t\)	\(s\)	\(p\)	\(u\)	\(r\)	\(q\)
\(u\)	\(u\)	\(p\)	\(s\)	\(t\)	\(q\)	\(r\)

Mark scheme (a)

Scheme

Marks

\(\times_{14}\)	1	3	5	9	11	13
1	1	3	5	9	11	13
3	3	9	1	13	5	11
5	5	1	11	3	13	9
9	9	13	3	11	1	5
11	11	5	13	1	9	3
13	13	11	9	5	3	1

M1
A1
A1

1.1b
1.1b
1.1b

(3)

Notes

M1: At least 3 correct entries.

A1: At least 6 correct entries.

A1: All 12 entries correct.

NB for the final A all entries must be given modulo 14. Do not accept e.g. 17 for 3.

Mark scheme (b)

Scheme	Marks	AO
\(\{1, 13\}\)	B1	2.2a
	(1)

Notes

B1: Deduces the correct subgroup. This is the only subgroup of order 2. Accept with set brackets or round brackets as long as both elements are listed. Accept \(\langle 13 \rangle\) as long as the angled brackets are clear.

Mark scheme (c)

Scheme	Marks	AO
States the groups are cyclic or identifies a generator for each. E.g. \(3^2 = 9, 3^3 = 13, 3^4 = 11, 3^5 = 5, 3^6 = 1\) so 3 generates G or \(q^2 = t, q^3 = s, q^4 = r, q^5 = u, q^6 = p\) so \(q\) generates \(H\)	M1	3.1a
E.g. \((3^2 = 9, 3^3 = 13, 3^4 = 11, 3^5 = 5, 3^6 = 1)\) 3 (or 5) generates G and \((q^2 = t, q^3 = s, q^4 = r, q^5 = u, q^6 = p)\) \(q\) (or u) generates \(H\)	A1	1.1b
Both groups are cyclic of order 6 and so are isomorphic.	A1	2.1
	(3)
	(7 marks)

Notes

M1: State the groups are cyclic, or identifies a generator for each group.

A1: Correctly generator identified for each group.

A1: Correct work for showing a generator for each group and conclude both cyclic and hence isomorphic.

Alt I

Scheme

Marks

\(G\)	1	3	5	9	11	13
\(H\)	\(p\)	\(q\)	\(u\)	\(t\)	\(r\)	\(s\)
Or \(H\)	\(p\)	\(u\)	\(q\)	\(r\)	\(t\)	\(s\)

M1
A1

3.1a
1.1b

Also \(3^2 = 9\) and \(\theta(3)^2 = q^2 = t = \theta(9) = \theta(3^2)\) so structure is preserved and therefore groups are isomorphic

2.1

(3)

M1: Attempts to identify an isomorphism between the elements – may be implied by

identifying at least 4 correct pairings
by attempting to rearrange group tables to have the same structure

A1: For a correct pairing of all items for an isomorphism. Note that \(3^2 = 9\) while \(q^2 = t\) and this can be used to see if they have a fully correct pairing (check if \(3 \leftrightarrow q\) then you need \(9 \leftrightarrow t\), or if \(3 \leftrightarrow u\) then you need \(9 \leftrightarrow r\))

A1: All correct pairings with demonstration of structure being preserved for at least one of the possible choices made, or via matching of group tables to show same structure, and concludes isomorphic.

NB there are two choices for isomorphism, \(1 \leftrightarrow p\) and \(13 \leftrightarrow s\) are forced, 3 and 5 must map to \(q\) and \(u\) either way round, but then 9 and 11 need to be correctly selected for the order of 3 and 5 – and this must be shown for fully marks. They may likely choose the pairings by considering orders (as in Alt II) but still need a reason for the final choice.

Alt II

Scheme

Marks

\(G\)	1	3	5	9	11	13
Order	identity	6	6	3	3	2

\(H\)	\(p\)	\(q\)	\(r\)	\(s\)	\(t\)	\(u\)
Order	identity	6	3	2	3	6

M1
A1

3.1a
1.1b

Since orders match and group is order 6 (so one of only two types) therefore the groups are isomorphic isomorphic

2.1

(3)

M1: Attempts to find the order for each element of a both groups – at least four correct for at least one of the groups.

A1: All orders correct for all elements of both groups.

A1: All orders correct, reason referring to small group order and therefore groups are isomorphic. Note if they go on to give an incorrect pairing it is wrong reasoning so A0.

Note: In general matching orders of all elements is a necessary but not sufficient requirement for an isomorphism (e.g. for groups of order 16 there are distinct isomorphism classes which have the same number of elements of the same orders). However, for small groups (up to order 15) with few isomorphism classes this criterion is sufficient but for a fully correct answer some reference to the group order must be given.

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.

AS June 2024 Q1

EdexcelCurrent spec9 marksGroups

(i) The table below is a Cayley table for the group \(G\) with operation \(\circ\)

\(\circ\)	\(a\)	\(b\)	\(c\)	\(d\)	\(e\)	\(f\)
\(a\)	\(d\)	\(c\)	\(b\)	\(a\)	\(f\)	\(e\)
\(b\)	\(e\)	\(f\)	\(a\)	\(b\)	\(c\)	\(d\)
\(c\)	\(f\)	\(e\)	\(d\)	\(c\)	\(b\)	\(a\)
\(d\)	\(a\)	\(b\)	\(c\)	\(d\)	\(e\)	\(f\)
\(e\)	\(b\)	\(a\)	\(f\)	\(e\)	\(d\)	\(c\)
\(f\)	\(c\)	\(d\)	\(e\)	\(f\)	\(a\)	\(b\)

(a) State which element is the identity of the group. (1)

(b) Determine the inverse of the element \((b \circ c)\) (2)

(d) Show that the set \(\{b, d, f\}\) is a subgroup of \(G\). (2)

(ii) Given that \(H\) is a group with an element \(x\) of order 3 and an element \(y\) of order 6 satisfying\[yx = xy^5\]show that \(y^3xy^3x^2\) is the identity element. (3)

Scheme	Marks	AO
Identity is \(d\)	B1	2.2a
	(1)

Mark scheme (i)(b)

Scheme	Marks	AO
\((b \circ c)^{-1} = a^{-1}\) Alt: \((b \circ c)^{-1} = c^{-1} \circ b^{-1} = c \circ f\)	M1	1.1a
\(= a\) Alt: \(= a\)	A1	1.1b
	(2)

Notes

M1: Correct method to find the inverse of \(b \circ c\). May find \(b \circ c\) first and identify its inverse, or may apply inverse property and find \(c \circ f\).

A1: Correct element, \(a\).

Mark scheme (i)(c)

Scheme	Marks	AO
The set cannot be a subgroup because e.g. \(G\) has order 6 and 4 does not divide 6 so by Lagrange’s Theorem, or the identity is not in the set, or closure fails as e.g. \(a \circ a = d\) is not in the set.	B1	2.4
	(1)

Notes

B1: Any correct reason given. See scheme for possibilities.

Mark scheme (i)(d)

Scheme

Marks

Table for \(\{b, d, f\}\) is

	\(b\)	\(d\)	\(f\)
\(b\)	\(f\)	\(b\)	\(d\)
\(d\)	\(b\)	\(d\)	\(f\)
\(f\)	\(d\)	\(f\)	\(b\)

2.1

This is closed, as no new elements, \(b\) and \(f\) are inverses and \(d\) is the identity, so the subset is a subgroup.

1.1b

(2)

Notes

M1: Investigates closure, e.g. by drawing the table for \(\{b, d, f\}\) or finding the individual products (at least 3) or by considering \(\langle b \rangle = \{b, b^2, b^3\} = \{b, f, d\}\) so cyclic.

A1: Correctly shows closure, and refers to inverse and identity, and makes conclusion it is a subgroup.

Mark scheme (ii)

Scheme	Marks	AO
\(y^3xy^3x^2 = y^2yxy^3x^2 = y^2xy^5y^3x^2\)	M1	1.1b
\(= yyxy^8x^2 = yxy^5y^8x^2 = xy^5y^{13}x^2\)	M1	3.1a
\(= xy^{18}x^2 = xex^2 = x^3 = e\) *	A1*	2.1
	(3)
	(9 marks)

Notes

M1: Applies the relation \(yx = xy^5\) at least once to change positions of an \(x\) and a \(y\). May work from either end.

M1: Completes the process of gathering all \(y\)’s or \(x\)’s together by using the relation repeatedly. May reduce elements according to order throughout rather than at the end.

A1*: Uses the orders of the elements to reduce gathered \(y\) terms and \(x\) terms down to identity to complete the proof.

A2 June 2023 Q7

EdexcelCurrent spec6 marksGroups

7. The set \(G = \mathbb{R} - \left\{-\dfrac{3}{2}\right\}\) with the operation of \(x \bullet y = 3(x + y + 1) + 2xy\) forms a group.

(a) Determine the identity element of this group. (2)

(b) Determine the inverse of a general element \(x\) in this group. (3)

Mark scheme (a)

Scheme	Marks	AO
\(x \bullet e = x \Rightarrow 3(x + e + 1) + 2xe = x\) \(\Rightarrow 3e + 2xe = x - 3x - 3 \Rightarrow e = \ldots\) Or Let \(e = -1 \Rightarrow e \bullet y = 3(-1 + y + 1) + 2(-1)y = \ldots\)	M1	2.1
\[\Rightarrow e = -\frac{2x + 3}{2x + 3} \Rightarrow e = -1\]Or \(e \bullet y = 3(-1 + y + 1) + 2(-1)y = y\) therefore \(e = -1\)	A1	2.2a
	(2)

Notes

M1: Sets up and solves a correct equation for the identity element.
Alternatively spots the identity is \(-1\) and attempts to show this satisfies the identity property.
E.g. \(-1 \bullet x = 3(-1 + x + 1) + 2(-1)x = \ldots\)

A1: Correct identity / correct demonstration that \(-1\) is the identity with appropriate conclusion. (Note only one side needs to be checked as we are told \(G\) is a group so one side is sufficient by uniqueness).

Special case: states \(e = -1\) with no working scores M1A0

Mark scheme (b)

Scheme	Marks	AO
Let the inverse of \(x\) be \(y\) then \(x \bullet y = e \Rightarrow 3(x + y + 1) + 2xy = -1\)	M1	3.1a
\(\Rightarrow 3y + 2xy = -1 - 3x - 3 \Rightarrow y = \ldots\)	M1	2.1
\[\Rightarrow x^{-1} = -\frac{3x + 4}{2x + 3} \text{ oe such as } -\frac{3}{2} + \frac{1}{4x + 6}\]	A1	2.2a
	(3)

Notes

M1: Sets up a correct equation for the inverse using their identity element. May use \(x^{-1}\) throughout, or may use \(y\) or other variable.

M1: Expands, rearranges and by factorising out the inverse element from two terms to find an expression for the inverse element.

A1: Correct inverse element. Accept equivalent forms.

Mark scheme (c)

Scheme	Marks	AO
E.g. for \(x = -\tfrac{3}{2}\) the identity property would fail since we cannot divide by \(2x + 3 = 0\) or shows \(x^{-1} = -\tfrac{-\tfrac{1}{2}}{0}\) so no inverse or can’t divide by 0 Or \(-\tfrac{3}{2} \bullet x = 3\left(-\tfrac{3}{2} + x + 1\right) - 3x = -\tfrac{3}{2}\) so the identity property fails as not unique as any \(x\) would be an identity for \(-\tfrac{3}{2}\).	B1	2.4
	(1)
	(6 marks)

Notes

B1: Explains that the identity property would fail for \(x = -\dfrac{3}{2}\). Alternatively accept a similar argument that \(x = -\dfrac{3}{2}\) could not have an inverse due to dividing by zero. May refer back to work in (b), if done there, without need to repeat it.
Just saying undefined is B0 without a reason why.

Note: Reasons saying closure or associativity would fail are incorrect so B0, as both of these properties are satisfied on all of \(\mathbb{R}\)

AS June 2023 Q1

EdexcelCurrent spec8 marksGroups

1. The operation \(*\) is defined on the set \(G = \{0, 1, 2, 3\}\) by

\[x * y \equiv x + y - 2xy \pmod{4}\]

(a) Complete the Cayley table below.

\(*\)	0	1	2	3
0
1
2
3

(2)

(b) Show that \(G\) is a group under the operation \(*\)
(You may assume the associative law is satisfied.) (3)

(d) State whether \(G\) is a cyclic group, giving a reason for your answer. (1)

Mark scheme (b)

Scheme	Marks	AO
Identity element is 0 and there is closure	M1	2.1
0 is the identity so is self-inverse 1, 2 and 3 are self-inverse	M1	2.5
Associative law is assumed so \(G\) forms a group	A1	1.1b
	(3)

Notes

M1: States closure and identifies 0 as the identity element.

M1: Finds inverses for each element.

A1: Scores both previous method marks and states that the associative law is satisfied therefore \(G\) is a group

Scheme	Marks	AO
0 has order 1 and 1, 2 and 3 have order 2	M1 A1	1.1b 1.1b
	(2)

Mark scheme (d)

Scheme	Marks	AO
There is no element with order 4 therefore \(G\) is not a cyclic group Every element is its own inverse therefore no group generator therefore \(G\) is not a cyclic group	B1	2.4
	(1)
	(8 marks)

Notes

B1: See scheme

AS June 2022 Q3

EdexcelCurrent spec9 marksGroups Number Theory

(i) Let \(G\) be a group of order 5 291 848
Without performing any division, use proof by contradiction to show that \(G\) cannot have a subgroup of order 11 (3)

(ii)

(a) Complete the following Cayley table for the set \(X = \{2, 4, 8, 14, 16, 22, 26, 28\}\) with the operation of multiplication modulo 30

\(\times_{30}\)	2	4	8	14	16	22	26	28
2	4	8	16	28	2	14	22	26
4	8		2			28	14
8	16	2			8			14
14	28		22	16		8	4
16	2	4		14	16
22	14		26			4	2	16
26	22	14		4				8
28	26		14		28		8

(b) Hence determine whether the set \(X\) with the operation of multiplication modulo 30 forms a group.
[You may assume multiplication modulo \(n\) is an associative operation.]

(6)

Mark scheme (i)

Scheme	Marks	AO
Suppose \(G\) has a subgroup of order 11, then (by Lagrange’s Theorem) 11 must divide 5291848	M1	2.1
But \(5 - 2 + 9 - 1 + 8 - 4 + 8 = 23\)	M1	1.1b
23 is not divisible by 11, hence 11 does not divide \(\|G\|\), which contradicts Lagrange’s Theorem. Hence there is no subgroup of order 11.	A1	2.4
	(3)

Notes

M1: Sets up the proof by stating or implying that if there is a subgroup of order 11 then by Lagrange’s Theorem 11 must divide 5291848. May not mention Lagrange’s Theorem at this stage. A formal assumption is not required as long as it is implicit.

M1: Applies the divisibility test for 11. Look for an attempt at the alternating sum being used.

A1: Alternating sum is 23, so derives a contradiction as 11 does not divide \(|G|\), and conclusion made. Use of Lagrange’s Theorem must be clear, though it need not be named.

Mark scheme (ii)(b)

Scheme

Marks

As the row and column for 16 repeat the borders, 16 is an identity element for \((X, \times_{30})\)

2.2a

Each element has an inverse as follows:

\(x\)	2	4	8	14	16	22	26	28
\(x^{-1}\)	8	4	2	14	16	28	26	22

1.1b

Since we know \(\times_{30}\) is associative and as there are no new elements in the table, so \((X, \times_{30})\) is closed, hence \((X, \times_{30})\) is a group.

2.4

(6)

(9 marks)

Notes

B1: Identifies 16 as the identity element. No reason needed.

B1: Identifies all inverses or gives reason why each element has an inverse (may refer to each row and column containing the identity once only and symmetrically about the diagonal).

B1: Refers to closure and associativity to deduce \((X, \times_{30})\) is a group.

SC Allow B0B0B1ft for deducing not a group with valid reason if identity or inverse checks fail.

A2 June 2022 Q1

EdexcelCurrent spec6 marksGroups

1. The group \(\mathrm{S}_4\) is the set of all possible permutations that can be performed on the four numbers 1, 2, 3 and 4, under the operation of composition.

For the group \(\mathrm{S}_4\)

(a) write down the identity element, (1)

(b) write down the inverse of the element \(a\), where\[a = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix}\] (1)

(c) demonstrate that the operation of composition is associative using the following elements\[a = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \quad b = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix} \quad \text{and } c = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\] (2)

(d) Explain why it is possible for the group \(\mathrm{S}_4\) to have a subgroup of order 4
You do not need to find such a subgroup. (2)

Scheme	Marks	AO
\(\{e =\}\ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix}\)	B1	1.1b
	(1)

Scheme	Marks	AO
\(\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 1 & 2 \end{pmatrix}\)	B1	1.1b
	(1)

Mark scheme (c)

Scheme	Marks	AO
Demonstrates that, for example: \([a \circ b] \circ c = \left[\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\right] \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\) \(= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}\) \(a \circ [b \circ c] = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \left[\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\right]\) \(= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}\)	M1	2.1
So \([a \circ b] \circ c = a \circ [b \circ c]\) or associative	A1	2.4
	(2)

Notes

M1: Shows two calculations in an attempt to show associative, e,g, \([a \circ b] \circ c\) and \(a \circ [b \circ c]\). There must be an intermediate line of working with evidence of using the permutations. Condone the wrong order for this mark.

A1: Correct calculations leading to \([a \circ b] \circ c = a \circ [b \circ c]\) or states associative

Note Incorrect order scores M1 A0

\[[a \circ b] \circ c = \left[\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\right] \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\]\[= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\]\[= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 2 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\]

Mark scheme (d)

Scheme	Marks	AO
The order of the group is 24 or 4!	B1	1.1b
4 is a factor of 24 or 4/24 therefore it is possible for a subgroup to have order 4.	B1ft	2.4
	(2)
	(6 marks)

Notes

B1: Order is 24 or 4!

B1ft: Follow through on their order of the group, draws the correct conclusion

A2 October 2021 Q4

EdexcelCurrent spec7 marksGroups Number Theory

4. Let \(G\) be a group of order \(46^{46} + 47^{47}\)

Using Fermat’s Little Theorem and explaining your reasoning, determine which of the following are possible orders for a subgroup of \(G\)

(i) 11

(ii) 21

(7)

Mark scheme (i)

Scheme	Marks	AO
(Order of a subgroup must divide the order of a group by Lagrange’s Theorem), so need to check if 11 (and/or 21) divides \(46^{46} + 47^{47}\) and by FLT, e.g. \(a^{11-1} = a^{10} \equiv 1 \pmod{11}\), so	M1	1.1b
\(46^{46} + 47^{47} \equiv 2^{4 \times 10 + 6} + 3^{4 \times 10 + 7} \equiv 2^6 + 3^7 \equiv 64 + (3^3)^2 \times 3\) \(\equiv 9 + 5^2 \times 3 \equiv 84 \equiv 7 \pmod{11}\)	M1	3.1a
Hence 11 is not a divisor of \(46^{46} + 47^{47}\) so not a possible order for a subgroup.	A1	2.2a

Notes

M1: For an attempt to apply a correct Fermat’s Little theorem at least once in the question with either \(p = 11\), \(p = 7\) or \(p = 3\) on either the \(46^{46}\) or \(47^{47}\) term.

M1: Applies FLT and congruence arithmetic fully to find the residue of \(46^{46} + 47^{47}\) modulo 11. There will be lots of different routes, so look for an attempt to apply FLT that leads to determining if 11 is a divisor or not.

A1: \(46^{46} + 47^{47} \equiv 7 \pmod{11}\) (accept equivalents as long as it is clear it is not congruent to 0) and deduces it is not a possible order for a subgroup.

Mark scheme (ii)

Scheme	Marks	AO
\(21 = 7 \times 3\) so need to check for factors of 7 and 3, using \(a^2 \equiv 1 \pmod{3}\) and \(a^6 \equiv 1 \pmod{7}\)	M1	3.1a
\(46^{46} + 47^{47} \equiv 1^{46} + 2^{47} \equiv 1 + 2^{2 \times 23 + 1} \equiv 1 + 2^1 \equiv 3 \equiv 0 \pmod{3}\)	M1	1.1b
\(46^{46} + 47^{47} \equiv 4^{46} + (-2)^{47} \equiv 4^{6 \times 7 + 4} + (-2)^{6 \times 7 + 5} \equiv 4^4 + (-2)^5\) \(\equiv 16^2 - 32 \equiv 9^2 - 4 \equiv 81 - 4 \equiv 77 \equiv 0 \pmod{7}\)	M1	2.1
As \(46^{46} + 47^{47}\) divisible by both 3 and 7 it is divisible by 21 and hence this is a possible order for a subgroup.	A1	2.4
	(7)
	(7 marks)

Notes

M1: Applies checks for both 7 and 3 as divisors of \(46^{46} + 47^{47}\) via similar strategy.

M1: Applies FLT with \(p = 3\) to find a smaller residue modulo 3. Other routes are possible.

M1: Applies FLT with \(p = 7\) to find a smaller residue modulo 7. Other routes are possible.

A1: Shows \(46^{46} + 47^{47}\) congruent to 0 modulo 3 and modulo 7, and deduces 21 divides \(46^{46} + 47^{47}\) hence it is a possible order for a subgroup.

Alt:

M1: Reduces the bases modulo 21 and applies a power reduction technique using congruences for at least one of the power of 46 or 47

M1: Reduces fully by congruence arithmetic either the \(46^{46}\) or \(47^{47}\) term.

M1: Reduces fully by congruence arithmetic both the \(46^{46}\) and \(47^{47}\) terms

A1: Shows \(46^{46} + 47^{47}\) congruent to 0 modulo 21, and deduces 21 divides \(46^{46} + 47^{47}\) hence it is a possible order for a subgroup.

A2 October 2021 Q2

EdexcelCurrent spec8 marksGroups

2. A binary operation \(\bigstar\) on the set of non-negative integers, \(\mathbb{Z}_0^+\), is defined by

\[m \mathbin{\bigstar} n = |m - n| \qquad m, n \in \mathbb{Z}_0^+\]

(a) Explain why \(\mathbb{Z}_0^+\) is closed under the operation \(\bigstar\) (1)

(b) Show that 0 is an identity for \((\mathbb{Z}_0^+, \bigstar)\) (2)

(d) Determine if \(\mathbb{Z}_0^+\) forms a group under \(\bigstar\), giving clear justification for your answer. (3)

Mark scheme (a)

Scheme	Marks	AO
For \(m, n \in \mathbb{Z}_0^+\) we have \(m - n \in \mathbb{Z}\) (difference of integers is an integer) and so \(\|m - n\| \in \mathbb{Z}_0^+\), hence closed under \(\star\).	B1	2.4
	(1)

Notes

B1: Checks difference of two non-negative integers is an integer and hence its modulus is a non-negative integer and concludes closure. “Always positive” as a conclusion is B0 without consideration of the equal zero case.

Mark scheme (b)

Scheme	Marks	AO
For \(m \in \mathbb{Z}_0^+,\ 0 \star m = \|0 - m\| = \|-m\| = m\)	M1	1.1b
and \(m \star 0 = \|m - 0\| = \|m\| = m\) Hence 0 is an identity*.	A1*	2.1
	(2)

Notes

M1: Checks that 0 is a left or a right identity. (Checks either side)

A1*: Checks 0 works both sides as an identity and makes conclusion it is an identity. (Checks both sides and makes conclusion.)

Mark scheme (c)

Scheme	Marks	AO
For \(m \in \mathbb{Z}_0^+\), we need \(\|m - n\| = 0 \Rightarrow n = \ldots\) or shows \(\|m - m\| = \|0\| = 0\)	M1	2.2a
As \(\|m - m\| = 0\) for all \(m \in \mathbb{Z}_0^+\) each \(m\) is self-inverse.	A1	2.1
	(2)

Notes

M1: Realises \(m\) must be its own inverse for each \(m\) – accept if just stated \(m\) is self-inverse with no proof, or if an attempt is made to show it is self-inverse, or for an attempt to solve \(|m - n| = 0\)

A1: Each element is self-inverse with a full proof given.

Mark scheme (d)

Scheme	Marks	AO
Checks associativity – ie evaluates \(m \star (n \star p)\) and \((m \star n) \star p\) with letter or numbers.	M1	1.2
E.g, \(1 \star (2 \star 3) = 1 \star \|2 - 3\| = 1 \star 1 = 0\) but \((1 \star 2) \star 3 = \|1 - 2\| \star 3 = 1 \star 3 = \|1 - 3\| = 2\)	M1	3.1a
\(1 \star (2 \star 3) \neq (1 \star 2) \star 3\) hence not associative so not a group.	A1	2.4
	(3)
	(8 marks)

Notes

M1: Realises associativity must be checked in some way – may be by producing a counter example, or by attempting to evaluate both sides of the associativity axiom for a general case. A statement of the correct identity is sufficient for the mark to be awarded.

M1: Produces a suitable counter example and evaluates both sides of associativity equation. Attempts at algebraic proofs are unlikely to succeed but allow the method for e.g consideration of. \(m \gt n \gt p\) giving \(\big||m - n| - p\big| = |m - n - p|\) and \(\big|m - |n - p|\big| = |m - n + p|\) but must have a correct reason to disambiguate the inner moduli. If in doubt use review.

A1: Must have provided a counter example. Deduces associativity does not hold and concludes \(\mathbb{Z}_0^+\) is not a group under \(\star\)

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).

Two example shadings of the star, each shading one half of every point in the same rotational sense (a pinwheel pattern), the second also with shaded spokes to the centre — Mark scheme (d)

AS October 2020 Q1

EdexcelCurrent spec8 marksGroups

1. The set \(G = \{1, 3, 7, 9, 11, 13, 17, 19\}\) under the binary operation of multiplication modulo 20 forms a group.

(a) Find the inverse of each element of \(G\). (3)

(b) Find the order of each element of \(G\). (3)

(d) Explain how the subgroup you found in part (c) satisfies Lagrange’s theorem. (1)

Mark scheme (a)

Scheme

Marks

1, 9, 11 and 19 are self-inverse

M1
A1

1.1b
1.1b

3	7	13	17
7	3	17	13

1.1b

(3)

Notes

M1: For any 2 of the self-inverse elements

A1: All 4 self-inverse elements correctly identified

B1: Correct inverses for the other elements

A2 June 2019 Q6

EdexcelCurrent spec12 marksGroups

(i) A binary operation \(*\) is defined on positive real numbers by\[a \mathbin{*} b = a + b + ab\]Prove that the operation \(*\) is associative. (4)

(ii) The set \(G = \{1, 2, 3, 4, 5, 6\}\) forms a group under the operation of multiplication modulo 7

(a) Show that \(G\) is cyclic. (2)

The set \(H = \{1, 5, 7, 11, 13, 17\}\) forms a group under the operation of multiplication modulo 18

(b) List all the subgroups of \(H\). (3)

Mark scheme (i)

Scheme	Marks	AO
\((a\mathbin{}b)\mathbin{}c = (a + b + ab)\mathbin{*}c = a + b + ab + c + (a + b + ab)c\)	M1	2.1
\(a\mathbin{}(b\mathbin{}c) = a\mathbin{*}(b + c + bc) = a + b + c + bc + a(b + c + bc)\)	M1	2.1
\(\underline{a + b + ab + c + (a + b + ab)c} = a + b + c + bc + ab + ac + abc\) \(\phantom{a + b + ab + c + (a + b + ab)c} = \underline{a + b + c + bc + a(b + c + bc)}\)	A1	2.2a
so \((a\mathbin{}b)\mathbin{}c = a\mathbin{}(b\mathbin{}c)\) which means \(*\) is associative	A1	2.4
	(4)

Notes

M1: Begins proof by correctly expanding \((a\mathbin{*}b)\mathbin{*}c\) or \(a\mathbin{*}(b\mathbin{*}c)\) to an expression in \(a\), \(b\) and \(c\). Note they may expand as \((a\mathbin{*}b)\mathbin{*}c = (a\mathbin{*}b) + c + (a\mathbin{*}b)c = a + b + ab + c + (a + b + ab)c\) which is equally fine.

M1: Makes progress towards the required result by attempting to expand both \((a\mathbin{*}b)\mathbin{*}c\) and \(a\mathbin{*}(b\mathbin{*}c)\), but be generous with the attempts for this method. May achieve this by working from left to right, so look for arriving at the other expression through a chain of equalities.

A1: For both underlined expressions (but accept eg. \(c(a + b + ab)\) for \((a + b + ab)c\)) and a correct expansion seen for each independently or part of a chain as shown. The expansion may have terms in different orders.

A1: Explains that \((a\mathbin{*}b)\mathbin{*}c = a\mathbin{*}(b\mathbin{*}c)\) means that \(*\) is associative. Depends on both M marks and a correct expression having been found.

Mark scheme (ii)(a)

Scheme	Marks	AO
\(3^2 = 2\quad 3^3 = 6\quad 3^4 = 4\quad 3^5 = 5\quad 3^6 = 1\) or \(5^2 = 4\quad 5^3 = 6\quad 5^4 = 2\quad 5^5 = 3\quad 5^6 = 1\) Or special case for M1A0 if powers not shown: 3 has order 6 so generates the group	M1	2.1
3 (or 5) has order 6 and so generates the group so \(G\) is cyclic	A1	2.4
	(2)

Notes

M1: Demonstrates understanding of the term cyclic by either attempting all the powers of 3 or 5. Accept for this a statement \(\langle 3\rangle = \{3, 2, 6, 4, 5, 1\}\) which shows the elements list in order of powers.

A1: Must have evaluated all powers of 3 or 5 correctly and explains why the group is cyclic. Accept as 3 generates the group, or as 3 has the same order of \(G\) as reason. Must refer to cyclic in conclusion.

Special case: Allow M1A0 for a correct explanation of why \(G\) is cyclic if the order of 3 (or 5) is stated as 6 without justification – but must include reference to either being a generator or having the same order as \(G\).

Mark scheme (ii)(b)

Scheme	Marks	AO
\(\{1\}\), \(H\)	B1	1.1b
\(\{1, 17\}\) or \(\{1, 7, 13\}\)	M1	1.1b
\(\{1, 17\}\) and \(\{1, 7, 13\}\) (and no others)	A1	1.1b
	(3)

Notes

(You may ignore references to the operation for this part)

B1: Identifies \(\{1\}\) and \(H\) as subgroups

M1: Identifies \(\{1, 17\}\) or \(\{1, 7, 13\}\) as a subgroup

A1: Identifies \(\{1, 17\}\) and \(\{1, 7, 13\}\) as subgroups and no others

Mark scheme (ii)(c)

Scheme

Marks

\(G\)	1	2	3	4	5	6
\(H\)	1	7	5	13	11	17

\(G\)	1	2	3	4	5	6
\(H\)	1	13	11	7	5	17

M1
A1
A1

3.1a
1.1b
1.1b

(3)

(12 marks)

Notes

M1: Attempts to identify an isomorphism between the groups – may be implied by

identifying at least 2 correct non-identity pairings or
by attempting to rearrange group tables to have the same structure, or
by attempting to map powers of a generator to powers of a generator e.g \((\text{their } 3)^k \to (\text{their } 5)^k\) or
by matching of non-trivial proper subgroups to each other.

A1: Identifies 4 correct pairings, or sets up a mapping with one correct generator

A1: All pairings correct, or sets up a mapping with generators of each group correct, eg. \(3^k \to 5^k\)

AS June 2019 Q4

EdexcelCurrent spec7 marksGroups

4. The set \(\{e, p, q, r, s\}\) forms a group, \(A\), under the operation \(*\)

Given that \(e\) is the identity element and that

\[p * p = s \qquad s * s = r \qquad p * p * p = q\]

(a) show that

(i) \(p * q = r\)

(ii) \(s * p = q\)

(2)

(b) Hence complete the Cayley table below.

\(*\)	\(e\)	\(p\)	\(q\)	\(r\)	\(s\)
\(e\)
\(p\)
\(q\)
\(r\)
\(s\)

(2)

A student states that there is a subgroup of \(A\) of order 3

(d) Comment on the validity of this statement, giving a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks	AO
\(p * q = p * p * p * p = s * s = r\) OR \(s * s = r \Rightarrow p * p * p * p = r \Rightarrow p * q = r\)	B1	2.1
\(s * p = p * p * p = q\) OR as \(p * p * p = q\) and \(p * p = s \Rightarrow s * p = q\)	B1	2.1
	(2)

Notes

B1: Correct proof to achieve the printed statement

Mark scheme (b)

Scheme

Marks

\(*\)	\(e\)	\(p\)	\(q\)	\(r\)	\(s\)
\(e\)	\(\boldsymbol{e}\)	\(\boldsymbol{p}\)	\(\boldsymbol{q}\)	\(\boldsymbol{r}\)	\(\boldsymbol{s}\)
\(p\)	\(\boldsymbol{p}\)	\(\boldsymbol{s}\)	\(\boldsymbol{r}\)	\(e\)	\(q\)
\(q\)	\(\boldsymbol{q}\)	\(r\)	\(p\)	\(s\)	\(e\)
\(r\)	\(\boldsymbol{r}\)	\(e\)	\(s\)	\(q\)	\(p\)
\(s\)	\(\boldsymbol{s}\)	\(\boldsymbol{q}\)	\(e\)	\(p\)	\(\boldsymbol{r}\)

M1
A1

1.1b
1.1b

(2)

Notes

(b) Marked B1 B1 on ePen

M1: Finds at least 13 correct entries – usually the highlighted

A1: Completely correct table

(The entries highlighted in the printed mark scheme are shown in bold.)

Mark scheme (d)

Scheme	Marks	AO
The order of a subgroup is a factor of the order of the group (Lagrange’s Theorem)	M1	1.2
As 3 is not a factor of 5, the student’s statement is wrong	A1	2.3
	(2)
	(7 marks)

Notes

M1: Some indication that the order of a subgroup must be a factor of the order of the group. May say that 3 is not a factor of 5 or equivalent

A1: Fully correct unambiguous statement that refers Lagrange’s theorem and either

3 is not a factor of 5
3 does not divide 5
5 is not divisible by 3

and comments that the student’s statement is incorrect. No contradictory statements

AS June 2018 Q2

EdexcelCurrent spec10 marksGroups

Figure 1: equilateral triangle ABC with A at the top, B bottom left and C bottom right; dashed lines x (through A, vertical), y (through B) and z (through C) meet at O — Figure 1

Figure 1 shows an equilateral triangle \(ABC\). The lines \(x\), \(y\) and \(z\) and their point of intersection, \(O\), are fixed in the plane. The triangle \(ABC\) is transformed about these fixed lines and the fixed point \(O\). The lines \(x\), \(y\) and \(z\) each pass through a vertex of the triangle and the midpoint of the opposite side.

The transformations \(I\), \(X\), \(Y\), \(Z\), \(R_1\) and \(R_2\) of the plane containing triangle \(ABC\) are defined as follows:

\(I\): Do nothing
\(X\): Reflect in the line \(x\)
\(Y\): Reflect in the line \(y\)
\(Z\): Reflect in the line \(z\)
\(R_1\): Rotate \(120^\circ\) anticlockwise about \(O\)
\(R_2\): Rotate \(240^\circ\) anticlockwise about \(O\)

The operation \(*\) is defined as ‘followed by’ on the set \(T = \{I, X, Y, Z, R_1, R_2\}\).
For example, \(X * Y\) means a reflection in the line \(x\) followed by a reflection in the line \(y\).

(a)

(i) Complete the Cayley table below

		Second transformation
	\(*\)	\(I\)	\(X\)	\(Y\)	\(Z\)	\(R_1\)	\(R_2\)
First Transformation	\(I\)
	\(X\)		\(I\)			\(Z\)
	\(Y\)
	\(Z\)
	\(R_1\)		\(Y\)
	\(R_2\)

Given that the associative law is satisfied,

(ii) show that \(T\) is a group under the operation \(*\)

(6)

(b) Show that the element \(R_2\) has order 3 (2)

(d) Write down the elements of a subgroup of \(T\) that has order 3 (1)

Mark scheme (a)

Scheme

Marks

(a)(i)

\(*\)	\(I\)	\(X\)	\(Y\)	\(Z\)	\(R_1\)	\(R_2\)
\(I\)	\(I\)	\(X\)	\(Y\)	\(Z\)	\(R_1\)	\(R_2\)
\(X\)	\(X\)
\(Y\)	\(Y\)
\(Z\)	\(Z\)
\(R_1\)	\(R_1\)
\(R_2\)	\(R_2\)

1.1b

\(*\)	\(I\)	\(X\)	\(Y\)	\(Z\)	\(R_1\)	\(R_2\)
\(I\)	\(I\)	\(X\)	\(Y\)	\(Z\)	\(R_1\)	\(R_2\)
\(X\)	\(X\)	\(I\)	\(R_2\)	\(R_1\)	\(Z\)	\(Y\)
\(Y\)	\(Y\)	\(R_1\)	\(I\)	\(R_2\)	\(X\)	\(Z\)
\(Z\)	\(Z\)	\(R_2\)	\(R_1\)	\(I\)	\(Y\)	\(X\)
\(R_1\)	\(R_1\)	\(Y\)	\(Z\)	\(X\)	\(R_2\)	\(I\)
\(R_2\)	\(R_2\)	\(Z\)	\(X\)	\(Y\)	\(I\)	\(R_1\)

B1
B1

1.1b
1.1b

(ii)

\(I\) is the identity and closure is shown (by the Cayley table)

2.1

\(X\), \(Y\) and \(Z\) are self-inverse, \(R_1\) and \(R_2\) are inverses, (\(I\) is the identity so is self-inverse)

2.5

(Associative law may be assumed) so \(T\) forms a group

1.1b

(6)

Notes

(a)(i)

B1: Begins completing the table by having at least the first row and first column correct

B1: Mostly correct – three rows or three columns correct (so demonstrates an understanding of using \(*\))

B1: Fully correct table

(a)(ii)

M1: States closure and identifies the identity as \(I\)

M1: States the inverse of each element (reference to the Identity not required here)

A1: Concludes that \(T\) is a group (must see a conclusion)

Special case: If the inverses are not stated explicitly but a statement such as “all elements have an inverse” is seen, score M1M1A0

Mark scheme (b)

Scheme	Marks	AO
\(R_2 * R_2 * R_2 = (R_2 * R_2) * R_2\) or \(R_2 * (R_2 * R_2) = R_1 * R_2\) or \(R_2 * R_1\)	M1	2.1
\(= I\) (the identity) so \(R_2\) has order 3	A1	2.2a
	(2)

Notes

M1: Clearly begins process to find \(R_2 * R_2 * R_2\) reaching \(R_1 * R_2\) or \(R_2 * R_1\)

A1: Gives answer as \(I\) states identity and deduces that the order is 3 or e.g. \((R_2)^3 = I\)

Mark scheme (c)

Scheme	Marks	AO
\(R_1\) (and \(R_2\)) have order 3, \(X\), \(Y\) and \(Z\) have order 2 so: There is no element of \(T\) that generates the group or There is no element of order 6	B1	2.4
	(1)

Notes

B1: Demonstrates an understanding of the term cyclic by referring to the order of \(R_1\), \(X\), \(Y\) and \(Z\) and makes a suitable conclusion