Given an infinite group which is finitely generated, is there a proper maximal normal subgroup?
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$\begingroup$ 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 SpeyerCommented Nov 4, 2009 at 14:29
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$\begingroup$ I think that the title of the question (if indeed not the body) suggests that the question is "does one exist". $\endgroup$– José Figueroa-O'FarrillCommented Nov 4, 2009 at 14:35
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$\begingroup$ Yes, Iam interested in lesser goal : does one exists. $\endgroup$– arun sCommented Nov 4, 2009 at 14:38
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$\begingroup$ 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 MeckesCommented Nov 4, 2009 at 14:40
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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 GeraschenkoCommented Nov 4, 2009 at 14:51
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5 Answers
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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). 61--64.
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), 120-127.
answered Nov 4, 2009 at 14:40
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5$\begingroup$ But I think {1} should count as nontrivial. Otherwise, any finite simple group is a counter-example. I think an interesting question would be "Does Higman's group have a maximal proper normal subgroup?" $\endgroup$ Commented Nov 4, 2009 at 14:47
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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), 120-127. $\endgroup$ Commented Nov 4, 2009 at 14:48
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$\begingroup$ Our comments crossed. So I change my answer to yes (see Neumann's paper). $\endgroup$ Commented Nov 4, 2009 at 14:49
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$\begingroup$ @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 $\endgroup$ Commented Nov 4, 2009 at 14:58
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2$\begingroup$ Nice! And Neumann's paper is online jlms.oxfordjournals.org/cgi/reprint/s1-12/2/120 . $\endgroup$ Commented Nov 4, 2009 at 15:16
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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 universal-algebraic 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 0-ary laws).
Write $G_0$ for the substructure generated by the set of such constants (the set of elements that are image of some given 0-ary 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, zero-ary 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 $A\neq nA$.]
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So many answers! I'm completely lost. The paper of "B.H. Neumann, "Some remarks on infinite groups", Journal London Math. Soc, 12 (1937), 120-127" stated results for the existence of maximal subgroups, not maximal normal subgroup. Is this existence question of nontrivial normal subgroup still unsolved?
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2$\begingroup$ A maximal subgroup M of G is either normal or self-normalizing. 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. $\endgroup$ Commented Nov 5, 2009 at 3:19
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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. $\endgroup$ Commented May 17, 2010 at 0:31
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1$\begingroup$ Neumann states the result for maximal subgroups, but his proof works equally well for maximal normal subgroups. $\endgroup$ Commented Jun 17, 2017 at 1:16
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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.
answered Nov 4, 2009 at 14:39
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$\begingroup$ Well, it's not equivalent -- a group with an infinite simple group as a proper quotient has a maximal normal subgroup. $\endgroup$ Commented Nov 4, 2009 at 16:57
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$\begingroup$ 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. $\endgroup$ Commented Nov 4, 2009 at 18:54
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Check out the Tarski monster. It is 2-generated and simple.
Unless I misunderstood your question and you exclude infinite simple groups altogether.
answered Nov 4, 2009 at 14:49
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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 finitely-generated simple group will do, trivially. In any case, nothing from the monster properties is required here. $\endgroup$ Commented Nov 5, 2009 at 18:43
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$\begingroup$ You are right, of course. I had been thinking about this group sometime before I saw the question, hence the unnecessarily complicated "counter"example! $\endgroup$ Commented Nov 5, 2009 at 20:58