# How many elements does it take to normally generate a group?

This is a terminology question (I should probably know this, but I don't). Given a group $G$, consider the minimal cardinality $nr(G)$ of a set $S \subset G$ such that $G$ is the normal closure of $S$: $G = \langle\langle S \rangle \rangle$ (nr is short for normal rank). In other words, how many elements in $G$ do we need to kill to produce the trivial group? What is this invariant called? "Corank" and "normal rank" seem to mean other things, and I'm not sure what other terms to search for.

Also, what methods are there to get a lower bound on $nr(G)$, say when $G$ is finitely generated? A trivial lower bound in this case is $rk(H_1(G))$, since clearly $nr(A)=rank(A)$ for $A$ a finitely generated abelian group. One has $nr(G)\leq rank(G)$, since it suffices to kill a generating set, and if $G\to H$ is a surjection, then $nr(G)\geq nr(H)$.

• I've removed the tag "group", since this was the only question with that tag, and it seemed superfluous when "group theory" was already there. If there's a particular reason for having it you are welcome to put it back though. – Zev Chonoles Feb 14 '11 at 0:59

## 4 Answers

According, for example, to the following paper by Gonzales-Acuna

http://www.jstor.org/pss/1971036

the smallest number of elements needed to normally generate a group $G$ is called the weight of $G$. This terminology is confirmed in the book

Algebraic invariants of links

by J. Hillman. I also confirm that the "corank" of $G$ usually denotes the largest rank of a free quotient of $G$.

• For those who aren't aware of it, see Johnson's paper jstor.org/pss/2043056 for an easier proof of Gonzales-Acuna's theorem. – Autumn Kent Feb 10 '11 at 0:46
• Ah, I read Johnson's paper, but I guess I should have read Gonzales-Acuna too! Thanks Roberto! – Ian Agol Feb 10 '11 at 4:39

If $G$ is residually $p$-finite or residually [locally indicable amenable], then the weight of $G$ is bounded below by the $b_1^{(2)}(G)+1$, where $b_1^{(2)}(G)$ denotes the first $\ell^2$-Betti number of $G$.

I conjecture that this is the case for all torsionfree groups, but I do not know how to prove this in general.

What you define as $nr(G)$ is indeed commonly referred to as the weight of a group, often written $w(G)$. In general, computing the weight of a finitely generated group, or even any sensible lower bounds on the weight, is very difficult. Wiegold posed the following problem in 1976:

Is it true that every finitely generated perfect group is the normal closure of one element? (i.e., has weight 1).

This can be found as problem 5.52 in the Kourovka notebook: http://arxiv.org/abs/1401.0300

Wiegold's question is `true' in the case of finite groups, and also solvable groups. See M. Chiodo, Finitely annihilated groups, Bull. Austral. Math. Soc. 90, No. 3, 404-417 (2014). In particular, Corollary 5.7 states:

Let $n > 1$ and let $G$ be a finite or solvable group. Then $w(G) = n$ if and only if $w(G^{ab}) = n$, and $w(G)=1$ if and only if $w(G^{ab})= 1$.

-Maurice

Although it doesn't directly answer your question, it seems interesting to consider groups in which every normal generating set is already a generating set.

At least in the finitely generated world, this is equivalent to "every maximal subgroup is normal", which is also equivalent to $G' \leq \Phi(G)$. For finite groups, this is equivalent to being nilpotent, but for finitely generated groups it may be a strictly weaker condition than nilpotence (I don't know an example however.)

In any case, such a group (for example, any f.g. nilpotent group) necessarily has $\mathrm{nr}(G) = \mathrm{rank}(G)$.