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Let $G$ be a finitely generated group. The weight $w(G)$ of $G$ is defined to be the minimum number of elements of $G$ whose normal closure in $G$ is the whole of $G$ (this is sometimes also called the normal rank). Obviously, $d(G^{\operatorname{ab}})\leq w(G) \leq d(G)$, where $d(G)$ is the rank of $G$.

A minimal generating set for $G$ does not necessarily contain a minimal normal generating set. The question is: does there always exist such a minimal generating set, i.e. one that realises the rank and contains a subset realising the weight?

If not, which conditions on $G$ would guarantee the existence of such a generating set?

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  • $\begingroup$ When you say "minimal generating set", do you require it to realize the rank? $\endgroup$ – YCor Jan 18 '14 at 0:58
  • $\begingroup$ @YvesCornulier Yes that is what I meant. I should have been clearer about that. $\endgroup$ – Tom Harris Jan 18 '14 at 12:34
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A silly special case is a weight 1 group which is 2-generator. Since it is weight 1, its abelianization is cyclic. For any pair of generators, one may perform Nielsen transformations to get a pair of generators so that one of the generators generates the abelianization. This generator then must also be a normal generator, since if we kill it, we get a 1-generator group which has trivial abelianization, namely the trivial group.

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    $\begingroup$ This shows more generally that is $w(G)\ge d(G)-1$ then the answer to the question is: yes, there exists a generating set of minimal size, containing a subset realizing the weight. $\endgroup$ – YCor Jan 18 '14 at 0:58
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The answer is positive for finitely generated soluble groups. (And also for finitely generated simple groups, but it's easy!)

Claim. Let $G$ be a finitely generated soluble group. Then $G$ has a generating set with $d(G)$ elements which contains a subset with $w(G)$ elements whose normal closure is $G$.

Proof. Let $W(G)$ be the intersection of the maximal normal subgroups of $G$. It follows from the definition of $W(G)$ that a vector in $G^n$ with $n \ge w(G)$ normally generates $G$ if and only if its image under the natural map $G \rightarrow \overline{G} \Doteq G/W(G)$ normally generates $\overline{G}$. Let $S$ be a generating vector of $G$ of length $d(G)$. By [1, Folgerung 2.10 and Satz 6.4], the group $\overline{G}$ is Abelian and we have $w(G) = w(G_{ab}) = w(\overline{G})$. Thus we can find a Nielsen transformation $\psi \in \text{Aut}(F_{d(G)})$ such that the last $d(G) - w(G)$ components of $S \cdot \psi$ lie in $W(G)$ [2, Theorem 1.1]. By the previous remark, this means that the first $w(G)$ components of $S \cdot \psi$ normally generate $G$.


[1] R. Baer, "Der reduzierte Rang einer Gruppe", 1964.
[2] D. Oancea, "A note on Nielsen equivalence in finitely generated abelian groups", 2011.

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