7 The group \(G\), of order 12, consists of the set \(\{ 1,2,4,5,8,10,13,16,17,19,20 , x \}\) under the operation of multiplication modulo 21 . The identity of \(G\) is the element 1 . The element \(x\) is an integer, \(0 < x < 21\), distinct from the other elements in the set.
An incomplete copy of the Cayley table for \(G\) is shown below:
| G | 1 | 2 | 4 | 5 | 8 | 10 | 13 | 16 | 17 | 19 | 20 | \(x\) |
| 1 | 1 | 2 | 4 | 5 | 8 | 10 | 13 | 16 | 17 | 19 | 20 | |
| 2 | 2 | 4 | 8 | 10 | 16 | 20 | 5 | \(x\) | 13 | 17 | 19 | |
| 4 | 4 | 8 | 16 | 20 | \(x\) | 19 | 10 | 1 | 5 | 13 | 17 | |
| 5 | 5 | 10 | 20 | 4 | 19 | 8 | 2 | 17 | 1 | \(x\) | 16 | |
| 8 | 8 | 16 | \(x\) | 19 | 1 | 17 | 20 | 2 | 10 | 5 | 13 | |
| 10 | 10 | 20 | 19 | 8 | 17 | 16 | 4 | 13 | 2 | 1 | \(x\) | |
| 13 | 13 | 5 | 10 | 2 | 20 | 4 | 1 | 19 | \(x\) | 16 | 8 | |
| 16 | 16 | \(x\) | 1 | 17 | 2 | 13 | 19 | 4 | 20 | 10 | 5 | |
| 17 | 17 | 13 | 5 | 1 | 10 | 2 | \(x\) | 20 | 16 | 8 | 4 | |
| 19 | 19 | 17 | 13 | \(x\) | 5 | 1 | 16 | 10 | 8 | 4 | 2 | |
| 20 | 20 | 19 | 17 | 16 | 13 | \(x\) | 8 | 5 | 4 | 2 | 1 | |
| \(x\) | | | | | | | | | | | | |
- State, with justification, the value of \(x\).
- In the table given in the Printed Answer Booklet, list the order of each of the non-identity elements of \(G\).
- Write down all the subgroups of \(G\) of order 3 .
- Write down all the subgroups of \(G\) of order 6 .
- Determine all the subgroups of \(G\) of order 4, and prove that there are no other subgroups of order 4.
- State, with a reason, whether \(G\) is a cyclic group.