6
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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
  • 6
    $\begingroup$ 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". $\endgroup$ – Henning Makholm Aug 5 '17 at 17:52
  • 1
    $\begingroup$ 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
7
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Groups where every proper subgroup is abelian, must be two generated. Infinite examples include things like Tarski monsters. $\endgroup$ – Paul Plummer Aug 5 '17 at 20:30
  • 1
    $\begingroup$ Thank you for supplying an example :) $\endgroup$ – mesel Aug 5 '17 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.