Every finitely generated group has simple quotient

Question

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

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
@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

score 2 · Accepted Answer · answered Jan 18 '18 at 18:12

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

Every finitely generated group has simple quotient

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