Just-infinite quotient in infinite finitely generated group

Question

Google gives partial access to the Handbook of Algebra by Michiel Hazewinkel. It is stated in this book that an infinite finitely generated group has a just-infinite quotient. I couldn't find a reference proving this result so I have tried myself... but I couldn't do it.

I have tried different things but mostly a start seeking for a contradiction and then using the third isomorphism theorem for groups and a construction from scratch of a just-infinite quotient. I guess my problem is I don't have much instinct here... Could anyone give me a hint? Thank you!!

Answer 1

The definition I have (from a paper by Caprace) for a just-infinite group adds the constraint that the group be infinite in addition to the condition that all quotients by non-trivial normal subgroups be finite.

The comments don't rule out the possibility of the maximal subgroup being of finite index, so the comments don't answer the question with my alternative definition.

Is the claim still true in the case of the alternative definition I've provided? In the paper I am reading, it is stated that the fact is a basic consequence of Zorn's lemma. I don't see how. What is the poset chosen in the application of Zorn's lemma? Normal subgroups of infinite index? It isn't clear to me how normal subgroups of infinite index have a maximal element.

Answer 2

The comments give an incomplete answer to this question.* I want to complete the answer now.

Definition. A group $K$ is just infinite if $K$ is infinite and every proper quotient of $K$ is finite.

Theorem. If a finitely generated group $G$ is infinite then $G$ has a just-infinite quotient.

Proof. The comments to the question point out that the result follows from the fact that $G$ contains some maximal normal subgroup of infinite index. We now prove that such a subgroup exists by Zorn's lemma. Consider the poset of infinite-index normal subgroups of $G$. Let $H_1\leq H_2\leq\ldots$ be a chain of infinite-index normal subgroups. To apply Zorn's lemma we require an infinite-index subgroup $H$, say, of $G$ such that for every element $H_i$ in our chain we have that $H_i<H$ (so $H$ is an upper bound). It then follows that the poset has a maximal element, and the result follows. So take $H:=\cup H_i$. This is a subgroup as if $a, b\in H$ then there exists some $i$ such that $a, b\in H_i$ and so $ab, a^{-1}\in H_i\leq H$ as required. It is therefore sufficient to prove that $H$ has infinite index in $G$. To see this, recall that $G$ is generated by some finite set, so if $H$ has finite index in $G$ then $H$ is also generated by a finite set $X$, say. There exists some subgroup $H_i$ such that $X\subset H_i$. Therefore $H_i$ has finite index in $G$, a contradiction. QED

I first saw this argument in a paper of Higman, where he constructed a finitely generated infinite group $G$ without any finite quotients. Applying the above argument to this group $G$ shows that there exists a finitely generated infinite simple group. Higman does not use Zorn's lemma. He instead cites a result of B.H.Neumann which proves, independently of Zorn's lemma, that every proper subgroup of a finitely generated group is contained in a maximal proper subgroup.

*The final comment (of freakish) currently reads "Actually the fact that every finitely generated group has a proper maximal subgroup is not trivial at all. Well, at least I don't remember how to prove it. Although I've seen the proof somewhere in some book".