I've been self-studying some Algebra, and recently did a set of exercises proving both that, for a surjective homomorphism <math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>&#x03D5;<!-- ϕ --></mi> </math> between two groups G and H, that if: (i) G is cyclic, then H is cyclic, and; (ii) G is abelian, then H is abelian.

Step 1
A nice thing to notice is that your choice of group H doesn't really matter - as long as you choose a non-cyclic or non-abelian group respectively, you can form a counterexample.
If we're being really lazy, we can just say "Let G be the trivial group and let $f:G\to <!--\; \to \; -->H$ be the map taking the identity to the identity" where the trivial group is the group that only has an identity - and is an example of a cyclic (and therefore abelian) group.
If we're being mildly lazy, we can also let G be whatever group we want and map it into any other group H by sending every element to the identity - this is called the zero map and can send any group to any other one, so there are truly no interesting things one can say simply because two groups have a homomorphism between them, because all pairs of groups do.If we're being a little less lazy and more insightful, we can notice that if H is any group, and x is any element thereof, then the set $\{\dots <!--\; \dots \; -->,x^{-<!--\; -\; -->2},x^{-<!--\; -\; -->1},e,x,x^2,\dots <!--\; \dots \; -->\}$, called the subgroup generated by x and noted by ⟨x⟩ is cyclic - and this is basically why cyclic groups are so important. However, ⟨x⟩ has an injective homomorphism into H: just send each element to itself. Thus, as long as you choose H to not have the properties you're asking about, this gives you maps from cyclic groups into it. If you want a more typical cyclic group, like Z/nZ, you can just choose x to have order n and send m∈Z/nZ to $x^m$ to create a homomorphism.
Step 2
A more abstract note, which is a theme one definitely learns well in algebra, is that surjective homomorphisms between objects tell you that the range is a quotient of the domain - meaning, basically, that the codomain just the domain with some things identified with each other. This often allows us to determine properties of the codomain from properties of the domain - as the exercise shows.
On the other hand, injective homomorphisms tell you that the domain is a subobject of the range - like how ⟨x⟩ is a subgroup of H. Since we only see a part of the whole picture, usually this means that we can deduce properties of the domain from properties of the codomain - so it works the other way around! (You can check that if $f:G\to <!--\; \to \; -->H$ is injective and H is abelian (or cyclic) then G is also abelian (or cyclic))