# Finite groups of which the centralizer of each element is normal.

Recently I noticed that if $G$ is a finite group and $g \in G$ for which the centralizer $C_G(g)$ is a normal subgroup, all of the elements of the conjugacy class $g^G$ commute with each other, and hence their product is a element of the center $Z(G)$ of $G$.

Now suppose that all of the centralizers of elements of $G$ are normal. Have these groups been classified? What can be said about these groups? I noticed that if $P$ is any Sylow $p$-subgroup of $G$ and $z \in Z(P)$, then $G=N_G(P)C_G(z)$ by the Frattini argument.

• If I recall correctly, these are the $2$-Engel groups. I don't have a reference, though, so I may be remembering wrong. – James May 31 '14 at 23:09
• James, you were right, see Mikko's answer. – Nicky Hekster Jun 1 '14 at 19:13
• A late response. This was observed by D. M. Rocke, $p$-groups with abelian centralizers, Proc. London Math. Soc. (1975), Prop. 3.6, as a consequence of Levi's theorem (the $p$-group condition was not used). – Siddhartha Jun 16 '18 at 14:59

I believe the comment by James is correct, these groups are precisely the $2$-Engel groups.

Claim: The following statements are equivalent for a group $G$.

1. Every centralizer in $G$ is a normal subgroup.

2. Any two conjugate elements in $G$ commute, ie. $x^g x = x x^g$ for all $x, g \in G$.

3. $G$ is a $2$-Engel group, ie. $[[x,g],g] = 1$ for all $x, g \in G$.

Proof:

1) implies 2): $x \in C_G(x)$, thus $x^g \in C_G(x)$ since $C_G(x)$ is normal.

2) implies 3): $x^g = x[x,g]$ commutes with $x$, thus $[x,g]$ also commutes with $x$.

3) implies 1): If $[[x,g],g] = 1$ for all $g \in G$, then according to Lemma 2.2 in [*], we have $[x, [g,h]] = [[x,g],h]^2$. Therefore $[C_G(x), G] \leq C_G(x)$, which means that $C_G(x)$ is a normal subgroup.

[*] Wolfgang Kappe, Die $A$-Norm einer Gruppe, Illinois J. Math. Volume 5, Issue 2 (1961), 187-197. link

• Hey Mikko, excellent!!' Thanks +1 from me. – Nicky Hekster Jun 1 '14 at 19:12

The first observation is that such a finite group is nilpotent. For, given any $x \in G,$ we certainly have $x \in F(C_{G}(x)),$ and the latter group is contained in $F(G)$ as $C_{G}(x) \lhd G.$ This essentially reduces the question to one about $p$-groups. Notice that any finite $p$-group $P$ of nilpotency class $2$ has the property (thanks to Jack Schmidt for pointing out an earlier error). I am not sure at the moment whether it is possible to characterize $p$-groups with the property in any straightforward fashion.