Given an infinite group which is finitely generated, is there a proper maximal normal subgroup?

$\begingroup$ Two trivial comments, not worth a full answer. (1) Obviously, the group itself is a maximal normal subgroup. I assume you want to exclude that. (2) The language "can one find" strikes me as a little ambiguous: I don't know whether you mean "does one exist?" or "is one computable/constructible?" I'm sure the answer to the second is "no", because almost everything in finitely presented groups is not computable. The first question is a nice problem. $\endgroup$– David E SpeyerNov 4, 2009 at 14:29

$\begingroup$ I think that the title of the question (if indeed not the body) suggests that the question is "does one exist". $\endgroup$– José FigueroaO'FarrillNov 4, 2009 at 14:35

$\begingroup$ Yes, Iam interested in lesser goal : does one exists. $\endgroup$– arun sNov 4, 2009 at 14:38

$\begingroup$ Regarding David Speyer's comment above, I think it's common practice to use "maximal" as shorthand "maximal proper" in many contexts. (For example, this is how Hungerford's Algebra uses it in "maximal ideal", etc.) $\endgroup$– Mark MeckesNov 4, 2009 at 14:40

2$\begingroup$ I assume the thing you're worried about is the union of a chain of proper normal subgroups being whole group. If you know that doesn't happen, you get existence of a maximal one by Zorn's lemma. $\endgroup$– Anton GeraschenkoNov 4, 2009 at 14:51
5 Answers
If you mean nontrivial maximal normal subgroup (not 1 or the whole group), then the answer is no.
Higman constructed a finitely generated infinite group $G$ with no subgroups of finite index. You then get a finitely generated group with no nontrivial normal subgroups by taking the quotient by a maximal normal subgroup.
Higman's group $G$ is $\langle a,b,c,d  a^{1} b a = b^2, b^{1}cb = c^2, c^{1}dc=d^2, d^{1}ad=a^2 \rangle$
See Higman, Graham. A finitely generated infinite simple group. J. London Math. Soc. 26, (1951). 6164.
Edit:
If you mean does it have a proper maximal normal subgroup, then the answer is yes:
Finitely generated groups have a (possibly trivial) maximal normal subgroup. Higman's reference for this is B.H. Neumann, "Some remarks on infinite groups ", Journal London Math. Soc, 12 (1937), 120127.

5$\begingroup$ But I think {1} should count as nontrivial. Otherwise, any finite simple group is a counterexample. I think an interesting question would be "Does Higman's group have a maximal proper normal subgroup?" $\endgroup$ Nov 4, 2009 at 14:47

5$\begingroup$ Note that this uses the fact that finitely generated groups have a (possibly trivial) maximal normal subgroup. Higman's reference for this is B.H. Neumann, "Some remarks on infinite groups ", Journal London Math. Soc, 12 (1937), 120127. $\endgroup$ Nov 4, 2009 at 14:48

$\begingroup$ Our comments crossed. So I change my answer to yes (see Neumann's paper). $\endgroup$ Nov 4, 2009 at 14:49

$\begingroup$ @arun s: Why "thank you anyway..."? Richard's first comment answers your question affirmatively. @Richard: you should edit that comment into the answer, since it's the "real" answer $\endgroup$ Nov 4, 2009 at 14:58

2$\begingroup$ Nice! And Neumann's paper is online jlms.oxfordjournals.org/cgi/reprint/s112/2/120 . $\endgroup$ Nov 4, 2009 at 15:16
In general, it is true for quite general structures: finitely generated ones have a maximal proper substructure as soon as they have at least proper substructure.
Namely consider universalalgebraic structure $G$: this means a set $G$ endowed with a given family of finitary laws $(\mu_i)_{i\in I}$. A substructure is a subset which is stable under every law (which in particular means it contains all constants, i.e., all images of 0ary laws).
Write $G_0$ for the substructure generated by the set of such constants (the set of elements that are image of some given 0ary law $\mu_i$). For instance, for a group or monoid $G$, we have $G_0=\{1_G\}$; for a semigroup $G$ we have $G_0=\emptyset$; for a unital ring $R$, we have $R_0=\mathbf{Z}1_R$, etc. Then $G$ has at least a proper substructure iff $G\neq G_0$.
The structure $G$ is said to be generated by a subset $Y$ the only substructure containing $Y$ is $G$; it is finitely generated if it is generated by some finite subset.
Now if $G$ is finitely generated (by some subset $Y$) and $G\neq G_0$, then Zorn applies to the set of proper substructures (indeed for a substructure $H$, the condition $H\neq G$ means that $H\cap Y\neq Y$, and this condition passes to increasing unions). Hence:
If $G$ is a finitely generated structure (with respect to this family of laws) and $G\neq G_0$ then $G$ has a maximal proper substructure.
This applies in particular to groups. For a group $G$, we have 3 laws (binary product, unary inverse, zeroary unit), so $G_0=\{1_G\}$.
Hence every finitely generated group $G\neq\{1\}$ has a maximal proper subgroup.
But given a fixed group $G$, this also applies to $G$groups: these are groups $H$ endowed with an action of $G$ (thus $I$ is formed of the 3 laws of the group $H$, and one unary law for each $g\in G$ defining the action of $G$, plus axioms that are not important here, saying that this is a $G$action. This applies to $G$ endowed to the $G$action by conjugation:
If $G\neq\{1\}$ is a group and $G$ is finitely generated as normal subgroup of itself (e.g., $G$ is finitely generated as group), then $G$ has a maximal proper normal subgroup.
[Remark: for an abelian group $A$, there is a maximal proper subgroup in $A$ iff $A$ is not divisible, i.e., iff there exists $n\ge 2$ such that $A\neq nA$.]
So many answers! I'm completely lost. The paper of "B.H. Neumann, "Some remarks on infinite groups", Journal London Math. Soc, 12 (1937), 120127" stated results for the existence of maximal subgroups, not maximal normal subgroup. Is this existence question of nontrivial normal subgroup still unsolved?

2$\begingroup$ A maximal subgroup M of G is either normal or selfnormalizing. In the latter case, the union U of a chain of normal subgroups contained in M will be normal and cannot equal G, as it lies in M. So U will be a maximal normal subgroup of G. $\endgroup$ Nov 5, 2009 at 3:19

3$\begingroup$ This argument is flawed. For example, $Sym(6)$ is a maximal subgroup of $Sym(7)$. The only normal subgroup of $Sym(7)$ which is contained in $Sym(6)$ is the identity subgroup, which clearly is not a maximal proper normal subgroup. $\endgroup$ May 17, 2010 at 0:31


1$\begingroup$ Neumann states the result for maximal subgroups, but his proof works equally well for maximal normal subgroups. $\endgroup$ Jun 17, 2017 at 1:16
Assuming you mean "does a maximal normal subgroup always exist?" (and that you don't care about computing it), here is a way to restate the problem. Notice that if G has no maximal normal subgroups, that means that every proper normal subgroup H of G is contained in a larger proper normal subgroup K of G. In particular, this means that the group G/H must not be finite; if it were, we could only find a finite chain of normal subgroups between H and G. So the question "does a maximal normal subgroup always exist" is the same as "must a finitely generated group have any finite nontrivial quotients?" I'm not sure what the answer to that is, but it seems like a useful restatement.

$\begingroup$ Well, it's not equivalent  a group with an infinite simple group as a proper quotient has a maximal normal subgroup. $\endgroup$ Nov 4, 2009 at 16:57

$\begingroup$ Yes, good point. What I meant to say is that if finitely generated groups always have at least one finite nontrivial quotient, then they must have a proper maximal subgroup. $\endgroup$ Nov 4, 2009 at 18:54
Check out the Tarski monster. It is 2generated and simple.
Unless I misunderstood your question and you exclude infinite simple groups altogether.

1$\begingroup$ Downvoted: If the "proper" in the initial question does include the trivial subgroup, then this example does not qualify. If it does not, than the question is not interesting as any finitelygenerated simple group will do, trivially. In any case, nothing from the monster properties is required here. $\endgroup$ Nov 5, 2009 at 18:43

$\begingroup$ You are right, of course. I had been thinking about this group sometime before I saw the question, hence the unnecessarily complicated "counter"example! $\endgroup$ Nov 5, 2009 at 20:58