Prove that all cyclic groups of the same order are isomorphic

Question

I want to prove that 2 cyclic groups of the same order are always isomorph. So, let $G,H$ be cyclic groups with $|G| = |H|$. Then: I suspect the isomorphism would be the one that maps a generator g to a generator h. Thus, consider this function:

$f: G \rightarrow H$ defined by $f(g^k) = h^k$. I already showed that this mapping is surjective and a homomorphism, but I get stuck at the injectivity part. Can someone give a hint (my attempts: I tried to deduce whether the kernel is $e_G$ and I also used the definition of injectivity but I got stuck in both approaches)?

Answer 1

Split into two cases, based on whether the groups are finite. For functions between finite sets of the same size (especially homomorphisms between finite groups of the same size), surjectivity implies bijectivity. For the infinite case, it should be easier to show that the kernel is trivial.

Answer 2

First proof based on cardinality: a surjective map between sets of same finite cardinality is injective.

Second proof based on the kernel and the first isomorphism theorem. The image of $f$, $f(G) = H$ is isomorphic to $G / \mathrm{ker} f$ and you get a contradiction if $\mathrm{ker} f \neq \{e_G\}$