I've been self-studying some Algebra, and recently did a set of exercises proving both that, for a surjective homomorphism $\phi$ 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.
Proving these two was simple enough, but now I'm thinking about removing the surjectivity constraint on $\phi$. It does not seem to me that these statements necessarily hold, in either case, but I am having some difficulty coming up with solid counterexamples to either.
I was considering using the Klein group as my $H$ group for part (i), and the dihedral group of order 3 ($D_3$) for part (ii), but I was getting stuck here, as I'm not sure precisely to what I should compare it in order to construct . Any hints/suggestions as to where to look would be greatly appreciated. Thanks.
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 $f:G\rightarrow H$ 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 $\{\ldots,x^{-2},x^{-1},e,x,x^2,\ldots\}$, called the subgroup generated by $x$ and noted by $\langle x\rangle$ is cyclic - and this is basically why cyclic groups are so important. However, $\langle x\rangle$ 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 $\mathbb Z/n\mathbb Z$, you can just choose $x$ to have order $n$ and send $m\in\mathbb Z/n\mathbb Z$ to $x^m$ to create a homomorphism.
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 $\langle x\rangle$ 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 $f:G\rightarrow H$ is injective and $H$ is abelian (or cyclic) then $G$ is also abelian (or cyclic))