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

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

2011-02-10T00:05:08Z

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 $nr(G)\geq nr(H)$.

Answer by Roberto Frigerio for How many elements does it take to normally generate a group?

2011-02-10T00:24:16Z

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 $G$.

Answer by Andreas Thom for How many elements does it take to normally generate a group?

2011-02-10T12:55:04Z

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.

Answer by ndkrempel for How many elements does it take to normally generate a group?

2011-02-14T00:41:03Z

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 $\mathrm{nr}(G) = \mathrm{rank}(G)$.