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.