- (i) Let \(G\) be a group of order 5291848
Without performing any division, use proof by contradiction to show that \(G\) cannot have a subgroup of order 11
(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 | |
A copy of this table is given on page 11 if you need to rewrite your Cayley table.
(b) Hence determine whether the set \(X\) with the operation of multiplication modulo 30 forms a group.
[0pt]
[You may assume multiplication modulo \(n\) is an associative operation.]
Only use this grid if you need to rewrite your Cayley table.
| \(\times _ { 30 }\) | 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 | |
(Total for Question 3 is 9 marks)