Set \(S = \{ {x_0},{\text{ }}{x_1},{\text{ }}{x_2},{\text{ }}{x_3},{\text{ }}{x_4},{\text{ }}{x_5}\} \) and a binary operation \( \circ \) on S is defined as \({x_i} \circ {x_j} = {x_k}\), where \(i + j \equiv k(\bmod 6)\).

(a)     (i)     Construct the Cayley table for \(\{ S,{\text{ }} \circ \} \) and hence show that it is a group.

  (ii)     Show that \(\{ S,{\text{ }} \circ \} \) is cyclic.

(b)     Let \(\{ G,{\text{ }} * \} \) be an Abelian group of order 6. The element \(a \in {\text{G}}\) has order 2 and the element \(b \in {\text{G}}\) has order 3.

  (i)     Write down the six elements of \(\{ G,{\text{ }} * \} \).

  (ii)     Find the order of \({\text{a}} * b\) and hence show that \(\{ G,{\text{ }} * \} \) is isomorphic to \(\{ S,{\text{ }} \circ \} \).

(a)     (i)     Cayley table for \(\{ S,{\text{ }} \circ \} \)

\(\begin{array}{*{20}{c|cccccc}}
   \circ &{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\
\hline
  {{x_0}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\
  {{x_1}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}} \\
  {{x_2}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}} \\
  {{x_3}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}} \\
  {{x_4}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}} \\
  {{x_5}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}
\end{array}\)     A4

Note: Award A4 for no errors, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.

S is closed under \( \circ \)     A1

\({x_0}\) is the identity     A1

\({x_0}\) and \({x_3}\) are self-inverses,     A1

\({x_2}\) and \({x_4}\) are mutual inverses and so are \({x_1}\) and \({x_5}\)     A1

modular addition is associative     A1

hence, \(\{ S,{\text{ }} \circ \} \) is a group     AG

(ii)     the order of \({x_1}\) (or \({x_5}\)) is 6, hence there exists a generator, and \(\{ S,{\text{ }} \circ \} \) is a cyclic group     A1R1

[11 marks]

(b)     (i)     e, a, b, ab     A1

and \({b^2},{\text{ }}a{b^2}\)     A1A1

Note: Accept \(ba\) and \({b^2}a\).

(ii)     \({(ab)^2} = {b^2}\)     M1A1

\({(ab)^3} = a\)     A1

\({(ab)^4} = b\)     A1

hence order is 6     A1

groups G and S have the same orders and both are cyclic     R1

hence isomorphic     AG

[9 marks]

Total [20 marks]

a) Most candidates had the correct Cayley table and were able to show successfully that the group axioms were satisfied. Some candidates, however, simply stated that an inverse exists for each element without stating the elements and their inverses. Most candidates were able to find a generator and hence show that the group is cyclic.

b) This part was answered less successfully by many candidates. Some failed to find all the elements. Some stated that the order of ab is 6 without showing any working.