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Let G and G1 be two groups such that G_{1} is a homomorphic image of G.

If G is a cyclic group then so is G_{1}.

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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)