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

Determine whether or not \(\{ H,{\text{ }} * \} \)  is a subgroup of \(\{ G,{\text{ }} * \} \).

closure: let A, B \( \in G\)

(because AB is a \(2 \times 2\) matrix)

and det(AB) = det(A)det(B) \( = 1 \times 1 = 1\)     M1A1

identity: the \(2 \times 2\) identity matrix has determinant 1     R1

inverse: let A \( \in G\). Then A has an inverse because it is non-singular     (R1)

since AA\(^{ - 1} = \) I, det(A)det(A\(^{ - 1}\)) = det(I) = 1 therefore A\(^{ - 1} \in G\)     R1

associativity is assumed

the four axioms are satisfied therefore \(\{ G,{\text{ }} * \} \) is a group     AG

[5 marks]

closure: let A, B \( \in H\). Then AB \( \in H\) because the arithmetic involved produces elements that are integers     R1

inverse: A\(^{ - 1} \in H\) because the calculation of the inverse involves interchanging the elements and dividing by the determinant which is 1     R1

the identity (and associativity) follow as above     R1

therefore \(\{ H,{\text{ }} * \} \) is a subgroup of \(\{ G,{\text{ }} * \} \)     A1

Note:     Award the A1 only if the first two R1 marks are awarded but not necessarily the third R1.

Note:     Accept subgroup test.

[4 marks]