The set \({{\rm{S}}_1} = \left\{ {2,4,6,8} \right\}\) and \({ \times _{10}}\) denotes multiplication modulo \(10\).

  (i)     Write down the Cayley table for \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) .

  (ii)     Show that \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) is a group.

  (iii)     Show that this group is cyclic.

Now consider the group \(\left\{ {{{\rm{S}}_1},{ \times _{20}}} \right\}\) where \({{\rm{S}}_2} = \left\{ {1,9,11,19} \right\}\) and \({{ \times _{20}}}\) denotes multiplication modulo \(20\). Giving a reason, state whether or not \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) and \(\left\{ {{{\rm{S}}_1},{ \times _{20}}} \right\}\) are isomorphic.

(i)

     A2

Note: Award A1 for one error.

(ii)     closure: it is closed because no new elements are formed     A1

identity: \(6\) is the identity element    A1

inverses: \(4\) is self-inverse and (\(2\), \(8\)) form an inverse pair     A1

associativity: multiplication is associative     A1

the four group axioms are satisfied

(iii)     any valid reason, e.g.

\(2\) (or \(8\)) has order \(4\), or \(2\) (or \(8\)) is a generator     A2

[8 marks]

the groups are not isomorphic     A1

any valid reason, e.g. \({{\rm{S}}_2}\) is not cyclic or all its elements are self-inverse     R2

[3 marks]

Parts (a) (i) and (a) (iii) were well answered in general. However, in (a) (ii), some candidates lost marks by not showing convincingly that \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) was a group. For example, in verifying the group axioms, some candidates just made bald statements such as "\(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) is closed". This was not convincing because the question indicated that it was a group so that closure was implied by the question. It was necessary here to make some reference to the Cayley table which showed that no new elements were formed by the binary operation. To gain full marks on this style of question candidates need to clearly explain the reasoning used for deductions.

In (b), most candidates realised that the quickest way to establish isomorphism (or not) was to determine the order of each element. Candidates who knew that there are essentially only two different groups of order four had a slight advantage in this question.