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]