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Date May 2017 Marks available 3 Reference code 17M.2.hl.TZ0.2
Level HL only Paper 2 Time zone TZ0
Command term Determine Question number 2 Adapted from N/A

Question

Consider the special case in which \(G = \{ 1,{\text{ }}3,{\text{  }}4,{\text{ }}9,{\text{ }}10,{\text{ }}12\} ,{\text{ }}H = \{ 1,{\text{ }}12\} \) and \( * \) denotes multiplication modulo 13.

The group \(\{ G,{\text{ }} * \} \) has a subgroup \(\{ H,{\text{ }} * \} \). The relation \(R\) is defined such that for  \(x\), \(y \in G\), \(xRy\) if and only if \({x^{ - 1}} * y \in H\). Show that \(R\) is an equivalence relation.

[8]
a.

Show that 3\(R\)10.

[4]
b.i.

Determine the three equivalence classes.

[3]
b.ii.

Markscheme

Reflexive: \(xRx\)     (M1) 

because \({x^{ - 1}}x = {\text{e}} \in H\)     R1

therefore reflexive     AG

Symmetric: Let \(xRy\) so that \({x^{ - 1}}y \in H\)     M1

it follows that \({({x^{ - 1}}y)^{ - 1}} = {y^{ - 1}}x \in H \Rightarrow yRx\)     M1A1

therefore symmetric     AG

Transitive: Let \(xRy\) and \(yRz\) so that \({x^{ - 1}}y \in H\) and \({y^{ - 1}}z \in H\)     M1

it follows that \({x^{ - 1}}y{\text{ }}{y^{ - 1}}z = {x^{ - 1}}z \in H \Rightarrow xRz\)     M1A1

therefore transitive (therefore \(R\) is an equivalence relation on the set  \(G\))     AG

[8 marks]

a.

attempt at inverse of 3: since \(3 \times 9 = 27 = 1(\bmod 13)\)     (M1)

it follows that \({3^{ - 1}} = 9\)     A1

since \(9 \times 10 = 90 = 12(\bmod 13) \in H\)     M1A1

it follows that 3\(R\)10     AG

[??? marks]

b.i.

the three equivalence classes are \(\{ 3,{\text{ }}10\} ,{\text{ }}\{ 1,{\text{ }}12\} \) and \(\{ 4,{\text{ }}9\} \)     A1A1A1

[??? marks]

b.ii.

Examiners report

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

Syllabus sections

Topic 4 - Sets, relations and groups » 4.2

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