Fact that GiGi 1 is cyclic of prime order to show that fGif

Fact that G_i/G_i + 1 is cyclic of prime order to show that f(G_i)/f(G_i+1) is either {1} or cyclic of prime order. Hence, conclude that there exist a chain of normal subgroups K = K_0 K_1 ... K_L = {1} with K_i/K_i + 1 cyclic of prime order.

Solution

f(G_i)/f(G_i+1) , being the image of f(G_i)/f(G_i+1) (f is surjective) has order dividing p.

Hence it is either trivial {1} or the cyclic group of order p.

Lifting this to the group G., the claim on the chain of normal subgroups follows