Copy and complete the following operation table.

(i)     Show that {S , \( * \)} is a group.

(ii)     Find the order of each element of {S , \( * \)}.

(iii)     Hence show that {S , \( * \)} is cyclic and find all the generators.

The set T is defined by \(\{ x * x:x \in S\} \). Show that {T , \( * \)} is a subgroup of {S , \( * \)}.

     A4

Note: Award A4 for all correct, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.

[4 marks]

(i)     closure: there are no new elements in the table     A1

identity: 8 is the identity element     A1

inverse: every element has an inverse because there is an 8 in every row and column     A1

associativity: (modulo) multiplication is associative     A1

therefore {S , \( * \)} is a group     AG

(ii)     the orders of the elements are as follows

     A4

Note: Award A4 for all correct, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.

(iii)     EITHER

the group is cyclic because there are elements of order 6     R1

OR

the group is cyclic because there are generators     R1

THEN

10 and 12 are the generators     A1A1 

[11 marks]

looking at the Cayley table, we see that

T = {2, 4, 8}     A1

this is a subgroup because it contains the identity element 8, no new elements are formed and 2 and 4 form an inverse pair     R2

Note: Award R1 for any two conditions

[3 marks]

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.