(i)     Write down the Cayley table for \(\{ G,{\text{ }}{ \times _7}\} \) .

(ii)     Determine whether or not \(\{ G,{\text{ }}{ \times _7}\} \) is cyclic.

(iii)     Find the subgroup of G of order 3, denoting it by H .

(iv)     Identify the element of order 2 in G and find its coset with respect to H .

The group \(\{ K,{\text{ }} \circ \} \) is defined on the six permutations of the integers 1, 2, 3 and \( \circ \) denotes composition of permutations.

(i)     Show that \(\{ K,{\text{ }} \circ \} \) is non-Abelian.

(ii)     Giving a reason, state whether or not \(\{ G,{\text{ }}{ \times _7}\} \) and \(\{ K,{\text{ }} \circ \} \) are isomorphic.

(i)     the Cayley table is

     A3

Note: Deduct 1 mark for each error up to a maximum of 3.

(ii)     by considering powers of elements,     (M1)

it follows that 3 (or 5) is of order 6     A1

so the group is cyclic     A1

(iii)     we see that 2 and 4 are of order 3 so the subgroup of order 3 is {1, 2, 4}     M1A1

(iv)     the element of order 2 is 6     A1

the coset is {3, 5, 6}     A1

[10 marks]

(i)     consider for example

\(\left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&1&3
\end{array}} \right) \circ \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&3&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  1&3&2
\end{array}} \right)\)     M1A1

\(\left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&3&1
\end{array}} \right) \circ \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&1&3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  3&2&1
\end{array}} \right)\)     M1A1

Note: Award M1A1M1A0 if both compositions are done in the wrong order.

Note: Award M1A1M0A0 if the two compositions give the same result, if no further attempt is made to find two permutations which are not commutative.

these are different so the group is not Abelian     R1AG

(ii)     they are not isomorphic because \(\{ G,{\text{ }}{ \times _7}\} \) is Abelian and \(\{ K,{\text{ }} \circ \} \) is not     R1

[6 marks]