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.)

  • $\begingroup$ If every quotient group is abelian, then in particular $G/1$ is abelian, so $G$ is abelian. Perhaps you'd like to modify the question to what you call it when every nontrivial quotient is abelian? $\endgroup$ – Alex Ortiz Aug 5 '17 at 17:42
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    $\begingroup$ You already have a simple "if and only if" condition. What could you conceivably want more than that -- anything else that works will be equivalent to "$N$ contains the commutator subgroup". $\endgroup$ – hmakholm left over Monica Aug 5 '17 at 17:52
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    $\begingroup$ note that the idea behind the statement you found is that we "kill" every "bad" term of the form $aba^{-1}b^{-1}$ in a non-formal way we are saying that $aba^{-1}b^{-1}=e$ or equivalently that $ab=ba$ $\endgroup$ – Yanko Aug 5 '17 at 18:21

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.

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    $\begingroup$ Groups where every proper subgroup is abelian, must be two generated. Infinite examples include things like Tarski monsters. $\endgroup$ – user29123 Aug 5 '17 at 20:30
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    $\begingroup$ Thank you for supplying an example :) $\endgroup$ – mesel Aug 5 '17 at 20:34

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