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Prove that any isomorphism between two cyclic groups always maps every generator to a generator.

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Step 1: Show that any homomorphism from a cyclic group is determined by its image on a generator of the group.

Step 2: Show that the image of the homomorphism is generated by the image of that generator.

Step 3: What can you say about the kernel of such a homomorphism? What powers of the generator does it send to the identity?

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Suppose you have two cyclic groups $G$ and $G'$ and some isomorphism $\phi\colon G\to G'$. Since $\phi$ is surjective, for any $g'\in G'$ there exists some $g\in G$ such that

$$g'=\phi(g).$$

Try expressing $g$ and/or $g'$ in terms of the generators, and see what happens from there. The fact that $\phi$ is a homomorphism is particularly useful here.

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