# Generation in a group versus generation in its abelianization.

## 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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Do you have easy counterexamples for free or closed surface groups? – Matthew Stover Mar 1 '10 at 20:39
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. – HJRW Mar 1 '10 at 20:59
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. – Richard Kent Mar 1 '10 at 21:36
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." – Richard Kent Mar 1 '10 at 21:46
More to the point: perfect groups are examples of groups where every subset weakly generates (but they don't have to generate). – Richard Kent Mar 1 '10 at 21:52