Date | May 2008 | Marks available | 16 | Reference code | 08M.3srg.hl.TZ2.1 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ2 |
Command term | Draw, Find, and Show that | Question number | 1 | Adapted from | N/A |
Question
(a) Draw the Cayley table for the set of integers G = {0, 1, 2, 3, 4, 5} under addition modulo 6, +6.
(b) Show that {G, +6} is a group.
(c) Find the order of each element.
(d) Show that {G, +6} is cyclic and state its generators.
(e) Find a subgroup with three elements.
(f) Find the other proper subgroups of {G, +6}.
Markscheme
(a) A3
Note: Award A2 for 1 error, A1 for 2 errors and A0 for more than 2 errors.
[3 marks]
(b) The table is closed A1
Identity element is 0 A1
Each element has a unique inverse (0 appears exactly once in each row and column) A1
Addition mod 6 is associative A1
Hence {G, +6} forms a group AG
[4 marks]
(c) 0 has order 1 (0 = 0),
1 has order 6 (1 + 1 + 1 + 1 + 1 + 1 = 0),
2 has order 3 (2 + 2 + 2 = 0),
3 has order 2 (3 + 3 = 0),
4 has order 3 (4 + 4 + 4 = 0),
5 has order 6 (5 + 5 + 5 + 5 + 5 + 5 = 0). A3
Note: Award A2 for 1 error, A1 for 2 errors and A0 for more than 2 errors.
[3 marks]
(d) Since 1 and 5 are of order 6 (the same as the order of the group) every element can be written as sums of either 1 or 5. Hence the group is cyclic. R1
The generators are 1 and 5. A1
[2 marks]
(e) A subgroup of order 3 is ({0, 2, 4}, +6) A2
Note: Award A1 if only {0, 2, 4} is seen.
[2 marks]
(f) Other proper subgroups are ({0}+6), ({0, 3}+6) A1A1
Note: Award A1 if only {0}, {0, 3} is seen.
[2 marks]
Total [16 marks]
Examiners report
The table was well done as was showing its group properties. The order of the elements in (b) was done well except for the order of 0 which was often not given. Finding the generators did not seem difficult but correctly stating the subgroups was not often done. The notion of a ‘proper’ subgroup is not well known.