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Background

I have been spending a lot of time in my research with subsets of groups that are very close to being generating sets. To make this precise:

Let $G$ be a group. If a subset $S$ of $G$ projects onto a generating set of $G/[G,G]$, we say that $S$ weakly generates $G$. The following fact (see page 350 in this book for a proof) shows that weak generation in nilpotent groups is a strong condition.

Fact. Let $G$ be a nilpotent group. If $S$ weakly generates $G$, then $S$ generates $G$.

In light of this result, we ask the following question:

Does there exist a finitely presented but non-nilpotent group $G$ such that every weakly generating subset of $G$ generates $G$?

If we drop the condition "finitely presented" then the first Grigorchuk group suffices. I'd be pretty surprised if no finitely presented examples exist.

Surface groups and free groups

In response to Matt's question below: For the free group $F(a,b)$, the set $\{a[[a,b],a],b \}$ weakly generates but doesn't generate (you can show this directly using uniqueness of freely reduced word form in a free group). You can use this to show that any non-abelian closed surface group has subsets that weakly generate but don't generate. For instance, in the genus two case, suppose $G$ has the standard presentation with generators $a,b,c,d$ and relation $[a,b][c,d]$. Consider the set $S = ${$a,b[[b,c],b],c,d$}. If this set generates G, then it generates the image $G/N$, where $N$ is the normal subgroup generated by $a$ and $d$. This image is a free group generated by the images of $b$ and $c$. The set S projects to a set which does not generate.

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    $\begingroup$ Do you have easy counterexamples for free or closed surface groups? $\endgroup$ – Matthew Stover Mar 1 '10 at 20:39
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    $\begingroup$ Matt: the elements a and b[a,b] obviously weakly generate <a,b>. To see that they don't generate, we can compute the Whitehead graph, and see that it doesn't have a cut vertex. $\endgroup$ – HJRW Mar 1 '10 at 20:59
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    $\begingroup$ In an infinite non-cyclic simple group, or any perfect group for that matter, the trivial element weakly generates. So, I'm a little confused by the comments following the question. $\endgroup$ – Autumn Kent Mar 1 '10 at 21:36
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    $\begingroup$ I'm just pointing out that if you drop "residually nilpotent" from the question, then a perfect group is not an answer to the question. I.e. it doesn't "suffice." $\endgroup$ – Autumn Kent Mar 1 '10 at 21:46
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    $\begingroup$ More to the point: perfect groups are examples of groups where every subset weakly generates (but they don't have to generate). $\endgroup$ – Autumn Kent Mar 1 '10 at 21:52
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Here is a little explanation for why nilpotent groups are the only possible example for groups satisfying sufficient finiteness properties. Finitely presented is barely a finiteness property at all, so certainly non-nilpotent examples may exist.

There are no finite non-nilpotent groups such that every weakly generating set is a generating set. If G is such a group, then [G,G] is finite and the condition is simply that the elements of [G,G] can be removed from any generating set of G. This is simply the condition that [G,G] ≤ Φ(G), where Φ(G) is the intersection of the maximal subgroups of G, also known as the Frattini subgroup. A finite group is nilpotent if and only if [G,G] ≤ Φ(G) (and of course [G,G] ≤ Φ(G) for any nilpotent group, finite or not), so this explains why nilpotent groups are important for this property.

Groups G in which [G,G] is noetherian and every subgroup has the property that every weakly generating set is a generating set have a different name: these are the groups in which every subgroup satisfies a very weak form of subnormality, known as being "serial" (sort of like being a member of a composition series whose order type is a fairly arbitrary linearly ordered set). This is described in section 12.4 of Robinson's Course in the Theory of Groups.

Finitely generated soluble groups have well behaved Frattini subgroups, so that if [G,G] ≤ Φ(G), then G is nilpotent. Assuming [G,G] is noetherian, a finitely generated soluble group G in which every weakly generating set is a generating set is nilpotent. In particular, polycyclic groups will not provide any examples.

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The goal of my answer is to provide references to articles that studied the class of groups under consideration.

Let $\mathcal{C}$ be the class of the finitely generated groups in which every weakly generating subset is a generating subset. Then this class has been studied, to some extent, in [1] and [2] in the context of the Andrews-Curtis conjecture.

Indeed, the authors considered in [1] and [2] the class $\mathcal{MN}$ of finitely generated groups in which every normally generating subset is a generating subset (the notation $\mathcal{MN}$ is used here because the latter class coincides with the class of groups whose maximal subgroups are normal; this is also the class of groups $G$ for which $[G, G] \subset \text{Frattini}(G)$).

Clearly, $\mathcal{MN}$ contains $\mathcal{C}$.

If every lift in $G$ of a generating subset of $G/[G, G]$ normally generates $G$, then $G \in \mathcal{C}$ if and only if $G \in \mathcal{MN}$. Finitely generated soluble groups have this property.

Unfortunately, none of these references answer OP's question: the non-nilpotent examples given in [2] are the Grigorchuck groups and the Gupta-Sidki groups and they don't admit any finite presentation.

Still, I find these articles helpful for they certainly improve our understanding of $\mathcal{C}$. Some stability conditions for $\mathcal{MN}$ are established in [2] (the class $\mathcal{MN}$ is stable under direct product [Corollary 2.4, 2], it enjoys some stability under taking finite index subgroups [Proposition 2.3, 2]) and many examples of groups that don't belong to $\mathcal{MN}$ are given (groups with finitely many ends [Proposition 2.5, 2], groups containing a non-abelian free subgroup of finite index [Corollary 1, 2]).

There is also this remark, which wasn't mentioned yet, see [Corollary 2, 2]:

A finitely generated linear group lies in $\mathcal{MN}$ if and only if it is nilpotent.

In [1], the class $\mathcal{MN}$ is identified with the class of groups with zero recalcitrance (the recalcitrance of a finitely generated group, if finite, is an integer that measures how well the group complies to the generalized Andrews-Curtis conjecture). So, the class of $r$-recalcitrance groups with $r > 0$ naturally generalizes $\mathcal{MN}$.


[1] "Recalcitrance in groups", R. G. Burns et al., 1999.
[2] "The class $\mathcal{MN}$ of groups in which all maximal subgroups are normal", A. Myropolska, arXive:1509.08090 [math.GR], 2015.

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