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.
Show that 3\(R\)10.
Determine the three equivalence classes.
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]
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]
the three equivalence classes are \(\{ 3,{\text{ }}10\} ,{\text{ }}\{ 1,{\text{ }}12\} \) and \(\{ 4,{\text{ }}9\} \) A1A1A1
[??? marks]