Groups killed by centralizing one element

Question

What groups $G$ contains an element $g$ such that $G/(g\text{ is made central})=1$ (or equivalently $[g,G]=G$, where $[g,G]:=\langle[g,h]\colon h\in G\rangle$)?

A necessary condition is that $G$ is a perfect group. A sufficient condition is that $G$ is a nonabelian simple group. However, it seems that neither condition is complete.

I'm happy to assume $G$ is finitely generated and finitely presented (actually, I'm happy to assume $G$ is a $3$-manifold group). Other interesting necessary and sufficient conditions are also welcome.

Answer 1

I hope that I am correctly interpreting the intended question.

I will use $x$ instead of $g$ to denote the distinguished element with the property that $[G,x] = G$, assuming there is such an element, and I will assume that $[G,x]$ denotes the subgroup $\langle [g,x]: g \in G \rangle,$ which is the usual group-theoretic interpretation of that notation.

Isn't it then the case that we have $G = [G,x]$ if and only if both $G$ is perfect, and $G$ is generated by the conjugates of $x$?

For if $G$ is generated by the conjugates of $x$, we have $G = \langle x \rangle [G,x],$ since $x^{g} = x(x^{-1}x^{g}) \in x[G,x]$ (and $[G,x]$ is a normal subgroup of $G$ by standard commutator identities).

Hence if $G$ is generated by conjugates of $x$ and $[G,x]$ is a proper normal subgroup of $G$, then $G$ has a non-trivial cyclic homomorphic image, and is not perfect.

Thus if $G$ is perfect, and is generated by the conjugates of $x$, then $G = [G,x].$

On the other hand, if $G = [G,x],$ then certainly $G = [G,G],$ so $G$ is perfect. Also, $G = \langle x^{-1}x^{g}: g \in G \rangle,$ so certainly $G = \langle x^{g} : g \in G \rangle$, and $G$ is generated by the conjugates of $x$.

In conclusion, a perfect group $G$ satisfies $G = [G,x]$ for some $x \in G$ if and only if $G$ is generated by one of its conjugacy classes.

Answer 2

Here is a proof for finite perfect groups. Possibly this is already what Dave Benson had in mind.

Let $G$ be a finite perfect group. If $G \cong G_1 \times G_2$ for nontrivial $G_1$ and $G_2$ then by induction there exists $x_i \in G_i$ such that $G_i = [G_i, x_i]$ and $G = [G_1, x_1][G_2, x_2] = [G, (x_1, x_2)]$, so we may assume $G$ is not a direct product. If $G$ is simple then $G = [G, x]$ for any nontrivial $x \in G$, so we may also assume $G$ is not simple. Let $N \triangleleft G$ be a minimal normal subgroup. Then by induction there is some $xN \in G/N$ such that $G/N = [G/N, xN]$, i.e., $G = MN$ where $M = [G, x]$. We cannot have $M \cap N = 1$ since $G$ is not a direct product, so by minimality of $N$ we must have $M \cap N = N$ and therefore $G = M$.