Verified answer

a) Show that φ(m+n) = φ(m)φ(n).

φ(m+n) = T^(m+n) = T^m*T^n = φ(m)φ(n)

b) Yes. If you do a few lower power examples, you'll quickly see that

T^n =

(1 n)

(0 1)

c) No.

(2 1)

(3 2)

is in SL2(Z) because its determinant is 1. Yet it is not of the form T^n for any n in Z.

d) ker(φ) = {0} because this is the only element of Z that maps to the identity element of SL2(Z), namely

(1 0)

(0 1)

You could also argue that since it is one-to-one, the kernel is trivial (because if it mapped more than one element of Z to the identity matrix, it wouldn't be one-to-one).