When is a quotient group abelian? Every subgroup of an abelian group is normal, and every quotient of an abelian group is abelian. Also, a subgroup of a nonabelian group need not be normal, and a quotient of a nonabelian group need not be abelian.

Is there a simple set of (sufficient, necessary) conditions for a quotient of a nonabelian group to be abelian?

I found one here. Let $G$ be a group with commutator subgroup $G'$, let $N$ be a normal subgroup of $G$. Now $G/N$ is a abelian iff $G'$ is a subset of $N$.
Are there others?

Bonus question.
  A nonabelian group where every proper subgroup is normal is called Hamiltonian.
  What do you call a nonabelian group where every proper subgroup is abelian?
  What do you call a nonabelian group where every quotient group is abelian? (Then the commutator subgroup is a subset of every normal subgroup.)

 A: Let $G$ be a group. 


*

*$G$ is abelian if and only if the mapping $g\mapsto g^{-1}$ is an isomorphism on the group $G$.

*If $G$ is finite and every irreducible character is linear then $G$ is abelian.

*If $Aut(G)$ acts on the set $G-\{e\}$ transitively then $G$ is abelian.

*If $\mathbb Z_2$ acts by automorphism on a finite group $G$ fixed point freely then $G$ is abelian.

*Let $A$ act on $G$ by automorphism and the action of $A$ is faithful. Assume that $[G,A,A]=1$ then both $A$ and $[G,A]$ is abelian.


If $G$ is a finite group in which every proper subgroup is abelian then $G$ must be a solvable group. Its proof is not so easy with elementary tools. Thus such groups must be meta-abelian groups. As far as I remember it is not true for infinite groups. (Paul Plummer supplied an example in comments.)
If $G$ is a finite solvable group in which every nontrvial quotient is abelian then $G'$ is a minimal normal subgroup. Thus, $G'$ is isomorphic to $\mathbb Z_p \times\mathbb Z_p \ldots \times \mathbb Z_p $. Hence $G$ is meta-abelian group. In that case,  One can also observe that $Q\in Syl_q(G)$ for $p\neq q$, $Q$ is abelian. $S_3,A_4$ are examples of such groups. 
But such groups need not be solvable in general. For example consider the group $G=(A_5\times A_5)\rtimes \mathbb Z_2$ where the action of the involution is the map $(a,b)\mapsto (b,a)$ on $A_5\times A_5$.
I think you should ask a more specific question though it seems that your question is too general.
