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Date May 2018 Marks available 3 Reference code 18M.3srg.hl.TZ0.1
Level HL only Paper Paper 3 Sets, relations and groups Time zone TZ0
Command term Find Question number 1 Adapted from N/A

Question

The binary operation multiplication modulo 10, denoted by ×10, is defined on the set T = {2 , 4 , 6 , 8} and represented in the following Cayley table.

Show that {T, ×10} is a group. (You may assume associativity.)

[4]
a.

By making reference to the Cayley table, explain why T is Abelian.

[1]
b.

Find the order of each element of {T, ×10}.

[3]
c.i.

Hence show that {T, ×10} is cyclic and write down all its generators.

[3]
c.ii.

The binary operation multiplication modulo 10, denoted by ×10 , is defined on the set V = {1, 3 ,5 ,7 ,9}.

Show that {V, ×10} is not a group.

[2]
d.

Markscheme

closure: there are no new elements in the table      A1

identity: 6 is the identity element      A1

inverse: every element has an inverse because there is a 6 in every row and column (2−1 = 8, 4−1 = 4, 6−1 = 6, 8−1 = 2)      A1

we are given that (modulo) multiplication is associative      R1

so {T, ×10} is a group      AG

[4 marks]

a.

the Cayley table is symmetric (about the main diagonal)      R1

so T is Abelian      AG

[1 mark]

b.

considering powers of elements      (M1)

     A2

Note: Award A2 for all correct and A1 for one error.

[3 marks]

c.i.

EITHER

{T, ×10} is cyclic because there is an element of order 4      R1

Note: Accept “there are elements of order 4”.

OR

{T, ×10} is cyclic because there is generator      R1

Note: Accept “because there are generators”.

THEN

2 and 8 are generators      A1A1

[3 marks]

c.ii.

EITHER

considering singular elements      (M1)

5 has no inverse (5 ×10 a = 1, a∈V has no solution)      R1

OR

considering Cayley table for {V, ×10}

     M1

the Cayley table is not a Latin square (or equivalent)      R1

OR

considering cancellation law

eg, 5 ×109 = 5 ×10 1 = 5      M1

if {V, ×10} is a group the cancellation law gives 9 = 1      R1

OR

considering order of subgroups

eg, {1, 9} is a subgroup      M1

it is not possible to have a subgroup of order 2 for a group of order 5 (Lagrange’s theorem)      R1

THEN

so {V, ×10} is not a group     AG

[2 marks]

d.

Examiners report

[N/A]
a.
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b.
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c.i.
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c.ii.
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d.

Syllabus sections

Topic 8 - Option: Sets, relations and groups » 8.9 » The order of a group.

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