TITLE: How many elements does it take to normally generate a group?
QUESTION [19 upvotes]: 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 $nr(G)\geq nr(H)$.

REPLY [6 votes]: 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