Complete the following Cayley table

[5 marks]

(i)     State the inverse of each element.

(ii)     Determine the order of each element.

Write down the subgroups containing

(i)     $r$,

(ii)     $u$.

    (M1)A4

Note:     Award M1 for use of Latin square property and/or attempted multiplication, A1 for the first row or column, A1 for the squares of $q$, $r$ and $s$, then A2 for all correct.

(i)     ${p^{ - 1}} = p,{\text{ }}{q^{ - 1}} = q,{\text{ }}{r^{ - 1}} = r,{\text{ }}{s^{ - 1}} = s$     A1

${t^{ - 1}} = u,{\text{ }}{u^{ - 1}} = t$     A1

Note:     Allow FT from part (a) unless the working becomes simpler.

(ii)     using the table or direct multiplication     (M1)

the orders of $\{ p,{\text{ }}q,{\text{ }}r,{\text{ }}s,{\text{ }}t,{\text{ }}u\}$ are $\{ 1,{\text{ }}2,{\text{ }}2,{\text{ }}2,{\text{ }}3,{\text{ }}3\}$     A3

Note:     Award A1 for two, three or four correct, A2 for five correct.

[6 marks]

(i)     $\{ p,{\text{ }}r\} {\text{ }}\left( {{\text{and }}({S_3},{\text{ }} \circ )} \right)$     A1

(ii)     $\{ p,{\text{ }}u,{\text{ }}t\} {\text{ }}\left( {{\text{and }}({S_3},{\text{ }} \circ )} \right)$     A1

Note:     Award A0A1 if the identity has been omitted.

Award A0 in (i) or (ii) if an extra incorrect “subgroup” has been included.

[2 marks]

Total [13 marks]

The majority of candidates were able to complete the Cayley table correctly. Unfortunately, many wasted time and space, laboriously working out the missing entries in the table - the identity is $p$ and the elements $q$, $r$ and $s$ are clearly of order two, so 14 entries can be filled in without any calculation. A few candidates thought $t$ and $u$ had order two.

Generally well done. A few candidates were unaware of the definition of the order of an element.

Often well done. A few candidates stated extra, and therefore incorrect subgroups.