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?

  • $\begingroup$ 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

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.

  • 3
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.