# Groups killed by centralizing one element

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.

• What does $G/(g\text{ is central})=1$ mean? By $[g,G]=G$ do you mean $G=\{[g,h]:h\in G\}$? Commented Aug 7 at 6:18
• I would guess it would mean $\langle [g,h] : h \in G \rangle$, but this should be clarified. Commented Aug 7 at 7:18
• For a finite group, perfect is enough. But an infinite perfect group can be the union of its proper normal subgroups. Commented Aug 7 at 7:23
• @DaveBenson One example is the subgroup of $\mathrm{Sym}(\aleph_\omega)$ of all permutations with support strictly less than $\aleph_\omega$. Is there a finitely generated example? Commented Aug 7 at 10:18
• I don't really understand the intended question. But it is a famous open question (e.g., in Baumslag's list) to find a finitely generated perfect group $G$ such that $G$ is not normally generated by any single element. As already mentioned here, it is easy to check that no finite $G$ yields an example.
– YCor
Commented Aug 7 at 11:14

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.

• For reference, this is called a “weight one” group. mathoverflow.net/a/54965/1345 Commented Aug 14 at 17:12
• It is clear that the weight of a group is no more than the rank. Also, a 2-generator perfect group has weight one (kill one generator, the quotient is cyclic and perfect). Commented Aug 14 at 17:24

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 $$G = M$$.

• Actually I was just going to kill the intersection of the maximal normal subgroups. In a perfect finite group, the quotient is a direct product of simple groups, and this is generated by the conjugates of any element that is not the identity in any component. Commented Aug 7 at 10:43
• This viewpoint (for finite groups) reminds me somewhat of a (not directly relevant) Theorem of Gasch\"utz : a finite group $G$ has a faithful complex irreducible character if and only if the socle of $G$ is generated by a single $G$-conjugacy class. Commented Aug 7 at 11:24