Here it is given that G1 is a homomorphic image of G.
So, there exist a function f: G to G1.
Suppose G is a cyclic group.
Let G = <a> (where a is the generator of G).
Hence, f(a) belongs to G1.
Let, f(a) = b.
Now, we will prove that G1 = <b> (That is b is the generator of G1)
Assume an arbitrary element u belongs to G1.
So, there exist y belongs to G such that f(y) = u.
Since G is cyclic and a is the generator of G we have y = an
Then u = f(y) = f(an) = f(a)n (Definition of homorphism of groups) = bn
Hence, G1 = <b> and G1 is a cyclic group (proved)