 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/9/b/solution

Write

M (a, b) = (a 0 b 1)

M (a, b) M (c, d) = M (a c, a d + b) .

Since

a, c \neq = 0

, the product is again in

G

M (1, 0)

, and

M (a, b)^{- 1} = M (a^{- 1}, - b / a)

lies in

G

z \mapsto a z + b

M (a, 0) M (c, 0) = M (a c, 0)

, it contains

M (1, 0)

, and

M (a, 0)^{- 1} = M (a^{- 1}, 0)

. Thus

G^{'}

is a subgroup. Finally,

f (M (a, b) M (c, d)) = M (a c, 0) = M (a, 0) M (c, 0) = f (M (a, b)) f (M (c, d)),

so $f : G \to G^{'}$ is a surjective group homomorphism.

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