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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!!

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    What is a just-infinite quotient? – 2017-02-01
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    A group is just-infinite if all quotients are finite. – 2017-02-01
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    That does not seem to be true then. There are finitely generated infinite simple groups which obviously don't have non-trivial quotients at all. Was there an assumption that the group is abelian? – 2017-02-01
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    @freakish - such groups are themselves just-infinite, and as they are trivial quotients of themselves, fail to serve as counter-examples. – 2017-02-01
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    @PaulSinclair Well, not having a proper quotient sounds like a cheat to me. But fair enough. In that case every finitely generated group has a proper (i.e. not being whole group) maximal normal subgroup. Therefore the quotient is simple so here you go. – 2017-02-01
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    Fair enough, thanks to both of you for the clarifications, I was trying to find something much too complicated! – 2017-02-01
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    @user412175 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. – 2017-02-02

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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 infinite.

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

(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. Now, the above link is behind a paywall for me (even though I am a fully paid up member of the LMS, based at a UK institution!), but if I recall correctly Higman does not use Zorn's lemma. He instead cites a result of B.H.Neumann which proves, independently of Zorn's lemma, that above theorem: that every infinite group contains a maximal normal subgroup of infinite index. Can anyone verify this?)

*The final comment 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".

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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.

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    Your definition is correct. See my answer, below. You really should have asked this as a new question though, and linked to this question. Answering this question like you did meant that your question was less likely to be answered. – 2017-11-22
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    Ok, note taken about how I should operate on the site, thanks for the advice . Correct me if I'm wrong, but the upper bound you choose in your application of Zorn's lemma is not an element of the poset of infinite index subgroups of G. (G has finite index in G) – 2017-11-22
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    To apply Zorn's lemma you don't want to pick the upper bound to be in the poset - if you did that then this upper bound would be the maximal element you are seeking and no application of Zorn's lemma is needed... – 2017-11-23
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    You defninitely want the upper bound of to any chain to be contained inside your poset. An element of the poset being an upper bound to one chain doesn't imply it is a maximal element... (Picking an element outside the poset doesn't make sense to begin with, you haven't defined an order relation outside of the poset...) A simple counterexample is the poset of finite subsets of the integers, for example. An upper bound outside of the poset is Z itself. So by your argument the finite subsets have a maximal element, which is a finite subset, which doesn't make any sense. – 2017-11-24
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    Of course, my mistake! I've edited my answer: you need finite generation of $G$, and the trick is to union the subgroups in your chain. – 2017-11-24
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    Thanks for your reply. I'm still not sure this is enough. What you've shown in your edit is that G neq H, but you want to show that H has infinite index. – 2017-11-24
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    Gah! I've amended the argument again. It is a very minor edit - just suppose $H=\cup H_i$ has finite index in $G$, note that $H$ is finitely generated and then apply the same idea as before. – 2017-11-27
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    Ah I see! The implication that if $H$ is a subgroup of $G$ and $G$ is finitely generated $\Rightarrow H$ is finitely generated was the missing piece of the puzzle. Thanks so much! I did a little bit of looking around and this statement is known as [Schreier's Lemma](https://groupprops.subwiki.org/wiki/Schreier%27s_lemma) – 2017-11-27