(a)     Show that {1, −1, i, −i} forms a group of complex numbers G under multiplication.

(b)     Consider \(S = \{ e,{\text{ }}a,{\text{ }}b,{\text{ }}a * b\} \) under an associative operation \( * \) where e is the identity element. If \(a * a = b * b = e\) and \(a * b = b * a\) , show that

(i)     \(a * b * a = b\) ,

(ii)     \(a * b * a * b = e\) .

(c)     (i)     Write down the Cayley table for \(H = \{ S{\text{ , }} * \} \).

(ii)     Show that H is a group.

(iii)     Show that H is an Abelian group.

(d)     For the above groups, G and H , show that one is cyclic and write down why the other is not. Write down all the generators of the cyclic group.

(e)     Give a reason why G and H are not isomorphic.

(a)

see the Cayley table, (since there are no new elements) the set is closed     A1

1 is the identity element     A1

1 and –1 are self inverses and i and -i form an inverse pair, hence every element has an inverse     A1

multiplication is associative     A1

hence {1, –1, i, –i} form a group G under the operation of multiplication     AG

[4 marks]

 

(b)     (i)     aba = aab

= eb     A1

= b     AG

 

(ii)     abab = aabb

= ee     A1

= e     AG

[2 marks]

 

(c)     (i)

     A2

Note: Award A1 for 1 or 2 errors, A0 for more than 2.

 

(ii)     see the Cayley table, (since there are no new elements) the set is closed     A1

H has an identity element e     A1

all elements are self inverses, hence every element has an inverse     A1

the operation is associative as stated in the question

hence {e , a , b , ab} forms a group G under the operation \( * \)     AG

 

(iii)     since there is symmetry across the leading diagonal of the group table, the group is Abelian     A1

[6 marks]

 

(d)     consider the element i from the group G     (M1)

\({{\text{i}}^2} = - 1\)

\({{\text{i}}^3} = - {\text{i}}\)

\({{\text{i}}^4} = 1\)

thus i is a generator for G and hence G is a cyclic group     A1

–i is the other generator for G     A1

for the group H there is no generator as all the elements are self inverses     R1

[4 marks]

 

(e)     since one group is cyclic and the other group is not, they are not isomorphic     R1

[1 mark]

Total [17 marks]

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. A number of candidates did not understand the term “Abelian”. Many candidates understood the conditions for a group to be cyclic. Many candidates did not realise that the answer to part (e) was actually found in part (d), hence the reason for this part only being worth 1 mark. Overall, a number of fully correct solutions to this question were seen.