Here it is given that G_{1} is a homomorphic image of G.
So, there exist a function f: G to G_{1}.
Suppose G is a cyclic group.
Let G = <a> (where a is the generator of G).
Hence, f(a) belongs to G_{1}.
Let, f(a) = b.
Now, we will prove that G_{1} = <b> (That is b is the generator of G_{1})
Assume an arbitrary element u belongs to G_{1}.
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 = a^{n}
Then u = f(y) = f(a^{n}) = f(a)^{n} (Definition of homorphism of groups) = b^{n}
Hence, G_{1} = <b> and G_{1} is a cyclic group (proved)