Challenging +1.2 This is a structured group theory question with clear scaffolding. Part (a) requires finding x using closure (straightforward modular arithmetic). Parts (b-c) involve systematic enumeration of elements and subgroups using Lagrange's theorem. Part (d) requires more careful analysis but the structure guides students. While this is Further Maths content, the question is methodical rather than requiring deep insight, making it moderately above average difficulty.
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.