# Minimal normally generating subsets of minimal generating sets

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?

• When you say "minimal generating set", do you require it to realize the rank? – YCor Jan 18 '14 at 0:58
• @YvesCornulier Yes that is what I meant. I should have been clearer about that. – Tom Harris Jan 18 '14 at 12:34

## 2 Answers

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.

• 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. – YCor Jan 18 '14 at 0:58

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.