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Let \(f: G \rightarrow H\) be a group homomorphism onto \(H\). If \(G\) is a cyclic group, prove that \(H\) is also cyclic.

Short Answer

Expert verified
Given a group \(G\) that is cyclic and a group homomorphism \(f: G \rightarrow H\), the group \(H\) is also cyclic, as every element of \(H\) can be represented as \(h^n = (f(g))^n = f(g^n)\), a power of a single element, i.e., it is generated by a single element.

Step by step solution

01

Define Generators

Start by defining the generator for the cyclic group \(G\). Let's say \(G\) is generated by a single element \(g\). This means every element in \(G\) can be written in the form \(g^n\), for some integer \(n\).
02

Apply Group Homomorphism

Then apply \(f\) to both sides. We have \(f(g^n) = (f(g))^n\) on the right because \(f\) is a homomorphism and preserves the group operation, so applying \(f\) maintains the exponent.
03

Define H Generators

Now let's say \(h = f(g)\) is an element in \(H\). Each element of \(H\) is \(h^n = (f(g))^n = f(g^n)\), by the properties of a homomorphism and definition of our generator \(h\). This means \(H\) is generated by \(f(g)\) or \(h\), where \(f(g)\) is an element of \(H\).
04

Conclude H is Cyclic

Based on the previous step, we can conclude that \(H\) is a cyclic group, as it can be generated by a single element.

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