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I've read that every finitely generated group has a simple quotient. Is it obvious?

Tom
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    Well it's not true for the trivial group. Otherwise it is an application of Zorn's lemma: if $G$ is not simple then it has a nontrivial normal subgroup that does not contain one of the generators $x$. Consider a normal subgroup that is maximal subject to not containing $x$, etc. – Derek Holt Jan 18 '18 at 15:35
  • Thank you. I don't know how to conclude from this, but I wrote a solution. Is it what you had in mind? – Tom Jan 18 '18 at 18:13
  • Yes your solution is more or less what I had in mind. – Derek Holt Jan 18 '18 at 19:31
  • @DerekHolt Probably YCor would have been along soon to point out that Zorn’s Lemma isn’t needed. To save him the trouble, I’ll point out his comment on this question: https://math.stackexchange.com/questions/1324763/non-hopfian-groups-that-only-have-quotients-that-are-themselves-or-the-trivial – Jeremy Rickard Jan 18 '18 at 19:48
  • @JeremyRickard Thanks for the pointer! But I think the proof that finitely generated groups have maximal subgroups needs Zorn's Lamma. – Derek Holt Jan 18 '18 at 21:51
  • @DerekHolt Doesn’t the same (choice free) proof work? – Jeremy Rickard Jan 18 '18 at 22:08

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Let $X$ be a finite set of generators of $G$.

Let $n$ be the largest number such that there is a normal subgroup $H\subsetneqq G$ containing $n$ of these generators.

The set of all such $H$ satifies the hypothesis of Zorn's lemma, and has a maximal element $H$. Now if $H \subsetneqq K\trianglelefteq G$,then $\# K\cap X > n$ and $K = G$ by definition of $n$.

$G/H$ is a simple group

Tom
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