Copy and complete the following Cayley table for this binary operation.

Give one reason why \(\{ S,{\text{ }}{ \times _{14}}\} \) is not a group.

Show that a new set G can be formed by removing one of the elements of S such that \(\{ G,{\text{ }}{ \times _{14}}\} \) is a group.

Determine the order of each element of \(\{ G,{\text{ }}{ \times _{14}}\} \).

Find the proper subgroups of \(\{ G,{\text{ }}{ \times _{14}}\} \).

     A4

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

[4 marks]

any valid reason, for example     R1

not a Latin square

7 has no inverse

[1 mark]

delete 7 (so that G = {1, 3, 5, 9, 11, 13})     A1

closure – evident from the table     A1

associative because multiplication is associative     A1

the identity is 1     A1

13 is self-inverse, 3 and 5 form an inverse

pair and 9 and 11 form an inverse pair     A1

the four conditions are satisfied so that \(\{ G,{\text{ }}{ \times _{14}}\} \) is a group     AG

[5 marks]

     A4

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

[4 marks]

{1}

{1, 13}\(\,\,\,\,\,\){1, 9, 11}     A1A1

[2 marks]

There were no problems with parts (a), (b) and (d).

There were no problems with parts (a), (b) and (d).

There were no problems with parts (a), (b) and (d) but in part (c) candidates often failed to state that the set was associative under the operation because multiplication is associative. Likewise they often failed to list the inverses of each element simply stating that the identity was present in each row and column of the Cayley table.

The majority of candidates did not answer part (d) correctly and often simply listed all subsets of order 2 and 3 as subgroups.