(a)     Consider the set A = {1, 3, 5, 7} under the binary operation \( * \), where \( * \) denotes multiplication modulo 8.

  (i)     Write down the Cayley table for \(\{ A,{\text{ }} * \} \).

  (ii)     Show that \(\{ A,{\text{ }} * \} \) is a group.

  (iii)     Find all solutions to the equation \(3 * x * 7 = y\). Give your answers in the form \((x,{\text{ }}y)\).

(b)     Now consider the set B = {1, 3, 5, 7, 9} under the binary operation \( \otimes \), where \( \otimes \) denotes multiplication modulo 10. Show that \(\{ B,{\text{ }} \otimes \} \) is not a group.

(c)     Another set C can be formed by removing an element from B so that \(\{ C,{\text{ }} \otimes \} \) is a group.

  (i)     State which element has to be removed.

  (ii)     Determine whether or not \(\{ A,{\text{ }} * \} \) and \(\{ C,{\text{ }} \otimes \} \) are isomorphic.

(a)     (i)         A3

Note: Award A2 for 15 correct, A1 for 14 correct and A0 otherwise.

(ii)     it is a group because:

the table shows closure     A1

multiplication is associative     A1

it possesses an identity 1     A1

justifying that every element has an inverse e.g. all self-inverse     A1

(iii)     (since \( * \) is commutative, \(5 * x = y\))

so solutions are (1, 5), (3, 7), (5, 1), (7, 3)     A2

Notes: Award A1 for 3 correct and A0 otherwise.

Do not penalize extra incorrect solutions.

[9 marks]

(b)    

Note: It is not necessary to see the Cayley table.

a valid reason     R2

e.g. from the Cayley table the 5 row does not give a Latin square, or 5 does not have an inverse, so it cannot be a group

[2 marks]

(c)     (i)     remove the 5     A1

(ii)     they are not isomorphic because all elements in A are self-inverse this is not the case in C, (e.g. \(3 \otimes 3 = 9 \ne 1\))     R2

Note: Accept any valid reason.

[3 marks]

Total [14 marks]

Candidates are generally confident when dealing with a specific group and that was the situation again this year. Some candidates lost marks in (a)(ii) by not giving an adequate explanation for the truth of some of the group axioms, eg some wrote ‘every element has an inverse’. Since the question told the candidates that \(\{ A,{\text{ }} * )\) was a group, this had to be the case and the candidates were expected to justify their statement by noting that every element was self-inverse. Solutions to (c)(ii) were reasonably good in general, certainly better than solutions to questions involving isomorphisms set in previous years.