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Let G and G1 be two groups such that G1 is a homomorphic image of G.

If G is a cyclic group then so is G1.

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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)

by Loyal (9.6k points)
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