Consider the following Cayley table for the set G = {1, 3, 5, 7, 9, 11, 13, 15} under the operation \({ \times _{16}}\), where \({ \times _{16}}\) denotes multiplication modulo 16.

(i)     Find the values of a, b, c, d, e, f, g, h, i and j.

(ii)     Given that \({ \times _{16}}\) is associative, show that the set G, together with the operation \({ \times _{16}}\), forms a group.

The Cayley table for the set \(H = \{ e,{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }}{b_1},{\text{ }}{b_2},{\text{ }}{b_3},{\text{ }}{b_4}\} \) under the operation \( * \), is shown below.

(i)     Given that \( * \) is associative, show that H together with the operation \( * \) forms a group.

(ii)     Find two subgroups of order 4.

Show that \(\{ G,{\text{ }}{ \times _{16}}\} \) and \(\{ H,{\text{ }} * \} \) are not isomorphic.

Show that \(\{ H,{\text{ }} * \} \) is not cyclic.

(i)     \(a = 9,{\text{ }}b = 1,{\text{ }}c = 13,{\text{ }}d = 5,{\text{ }}e = 15,{\text{ }}f = 11,{\text{ }}g = 15,{\text{ }}h = 1,{\text{ }}i = 15,{\text{ }}j = 15\)     A3

Note: Award A2 for one or two errors,

A1 for three or four errors,

A0 for five or more errors.

(ii)     since the Cayley table only contains elements of the set G, then it is closed     A1

there is an identity element which is 1     A1

{3, 11} and {5, 13} are inverse pairs and all other elements are self inverse     A1

hence every element has an inverse     R1

Note: Award A0R0 if no justification given for every element having an inverse.

since the set is closed, has an identity element, every element has an inverse and it is associative, it is a group     AG

[7 marks]

(i)     since the Cayley table only contains elements of the set H, then it is closed     A1

there is an identity element which is e     A1

\(\{ {a_1},{\text{ }}{a_3}\} \) form an inverse pair and all other elements are self inverse     A1

hence every element has an inverse     R1

Note: Award A0R0 if no justification given for every element having an inverse.

since the set is closed, has an identity element, every element has an inverse and it is associative, it is a group     AG

(ii)     any 2 of \(\{ e,{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }}{a_3}\} ,{\text{ }}\{ e,{\text{ }}{a_2},{\text{ }}{b_1},{\text{ }}{b_2}\} ,{\text{ }}\{ e,{\text{ }}{a_2},{\text{ }}{b_3},{\text{ }}{b_4}\} \)     A2A2 

[8 marks]

the groups are not isomorphic because \(\{ H,{\text{ }} * \} \) has one inverse pair whereas \(\{ G,{\text{ }}{ \times _{16}}\} \) has two inverse pairs     A2

Note: Accept any other valid reason:

e.g. the fact that \(\{ G,{\text{ }}{ \times _{16}}\} \) is commutative and \(\{ H,{\text{ }} * \} \) is not.

[2 marks]

EITHER

a group is not cyclic if it has no generators     R1

for the group to have a generator there must be an element in the group of order eight     A1

since there is no element of order eight in the group, it is not cyclic     A1

OR

a group is not cyclic if it has no generators     R1

only possibilities are \({a_1}\), \({a_3}\) since all other elements are self inverse     A1

this is not possible since it is not possible to generate any of the “b” elements from the “a” elements – the elements \({a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }}{a_4}\) form a closed set     A1

[3 marks]

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.