7
$\begingroup$

A group is non-Hopfian provided it is isomorphic to a proper quotient. The classic, finitely presented, example of such a group is the Baumslag-Solitar group $$BS(2,3)= \langle x,t \mid t^{-1}x^2 t =x^3 \rangle \cong \langle x,t \mid t^{-1}x^2 t =x^3, x=(x^{-1}t^{-1}xt)^2 \rangle$$ and the projection map is not an isomorphism.

  1. Is there a non-Hopfian, non-Abelian, group with the property every quotient is either isomorphic to itself or the trivial group? (I figure there must be)

  2. This should probably be a separate question... How about (1) for finitely generated, without assuming the axiom of choice? Similarly, how about (1) for finitely presented, without choice?

Note that $BS(2,3)$ is not such a group since we can quotient out $x$ and have a group isomorphic to $\mathbb{Z}$. Generally, if we have a finitely generated group, presented with minimal number of generators, with the above property, and quotient out by one of the generators, the new group can be generated by one less generator, so the quotient must be trivial.

Relevant: This question asks for any example without the non-Abelian restriction, and is given an Abelian group, the Prüfer group $\mathbb{Z}[1/p]/\mathbb{Z}$, and explains why, with the axiom of choice, there are no finitely generated examples.

$\endgroup$
  • $\begingroup$ Finitely generated and finitely presented groups are countable, so it's unlikely to be the source of new counterexamples. $\endgroup$ – Asaf Karagila Jun 14 '15 at 9:59
  • 1
    $\begingroup$ I looked at the argument in the answer you link; and it seems completely choice free. Zorn's lemma is used to produce a maximal normal subgroup not containing generators. But you can instead enumerate the elements of the group, and by induction construct a maximal normal subgroup not containing elements of the generating set. Simply add those which you can add, one step at a time. The set is countable, so no choice is needed. I can give you other, heavier more "black boxy" arguments from set theory, but I don't see why they are needed. $\endgroup$ – Asaf Karagila Jun 14 '15 at 10:30
  • 1
    $\begingroup$ I'm not certain, but I think for a nonabelian example one may take the infinite permutational wreath product$$\cdots\wr A_5\wr A_5.$$ $\endgroup$ – H.Durham Jun 14 '15 at 11:39
  • 1
    $\begingroup$ Does it work then? The reason I am uncertain is that this example is from a conversation I had a while ago. I was told then that if we do it finitely many times, then the lattice of normal subgroups is totally ordered. Now the direct limit should be a candidate. $\endgroup$ – H.Durham Jun 14 '15 at 15:10
  • 3
    $\begingroup$ You don't need the axiom of choice to discard finitely generated examples. Claim: every finitely generated group $\neq 1$ has a simple quotient. Proof: given a finite generating subset of $G$, get an epimorphism $p$ from a f.g. free group $F$. Enumerate (explicitly) $F$ as $(g_n)$. Define normal subgroups: by induction, $N_0=\{1\}$ and $N_{i+1}$ is the normal subgroup generated by $N_i$ and $p(g_i)$ if this gives a proper subgroup, otherwise $N_{i+1}=N_i$. Then $G/\bigcup N_i$ is simple. $\endgroup$ – YCor Oct 25 '16 at 21:46
3
$\begingroup$

I have provided a non-abelian example as a reply to the original MathSE question (where abelian was not discarded). (Sorry, I now realize I should maybe have replied here.)

This answers 1. As regards 2, as I said in a comment, there are no finitely generated examples regardless of the axiom of choice.

$\endgroup$

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.