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I think there is exactly one homomorphic image of $\mathbb{Z}_{36}$ up to an isomorphism, for each divisor of the group's order: $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$, as the group $\mathbb{Z}_{36}$ being abelian implies that each of its subgroups is normal. Therefore, there exists a homomorphism $\mathbb{Z}_{36} \rightarrow \mathbb{Z}_{36}/\mathbb{Z}_{k}$ for each divisor $k$.

I don't know how to proceed from there. In particular:

  • How to give an example of such homomorphism for each of the divisors?
  • How to construct a general isomorphism between the images of homomorphisms possible for a divisor?
  • How to prove there aren't any more homomorphic images?
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    Don't forget about $9$ ;) – 2017-01-29
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    Haha, I checked it twice and still lost one! – 2017-01-29
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    I noticed one was missing because the [number of divisors](https://en.wikipedia.org/wiki/Divisor_function#Properties) of $36=2^2\times3^2$ equals $(2+1)(2+1)=9$, but even knowing that I had to check twice. – 2017-01-29

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The quotient maps you suggest are already examples of such homomorphisms. Though the subgroups of $\Bbb{Z}_{36}$ aren't of the form $\Bbb{Z}_k$, they are only isomorphic to $\Bbb{Z}_k$ for some $k$ dividing $36$.

To prove that there aren't any more homomorphic images, you might use the isomorphism theorem. This also tells you that if two homomorphisms from $\Bbb{Z}_{36}$ have the same kernel, then their images are isomorphic.

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Hint: The image of a cyclic group is cyclic. The image of finite group of order $n$ is a finite group of order $d$ with $d$ a divisor of $n$.