Finite groups with bounded centralizers

Question

Let $G$ be a finite group. For each $x\in G$, the centralizer $\mathbf{C}_G(x)$ must contain $\langle x\rangle$.

QUESTION: What are some interesting results of the following form:

Given some bound on $|\mathbf{C}_G(x):\langle x\rangle|$ for all $1\neq x\in G$, the structure of $G$ is constrained in some way.

I doubt such results are "important" in any sense of the word, but they would definitely be amusing.

I have obtained a classification in the extreme case:

Theorem: Let $G$ be a finite group which is such that for every $1\neq x\in G$, $\mathbf{C}_G(x)=\langle x\rangle$. Then:

1. For some prime $p$, $G\cong C_p$
2. For distinct primes $p<q$, where $q\equiv 1\pmod p$, $G\cong C_q\rtimes C_p$ (the $\rtimes$ is unambiguous here because there is only one possible non-trivial action)

Proof of theorem:

---

Lemma 1: If $A\subseteq G$ is an abelian subgroup, then it is cyclic of prime order.

Proof:

For each $1\neq x\in A$, $A\subseteq\mathbf{C}_G(x)=\langle x\rangle\subseteq A$. It follows that $A=\langle x\rangle$ for every $1\neq x\in A$. This can only happen if $A$ is cyclic of prime order. $\blacksquare$

---

Lemma 2: If $K\subseteq G$ is a nilpotent subgroup, then it is cyclic of prime order.

Proof:

Let $z\in\mathbf{Z}(K)$ be a nonidentity central element (which is sure to exist, by the nilpotence of $K$). Then $K\subseteq\mathbf{C}_G(z)=\langle z\rangle\subseteq K$, so $K=\langle z\rangle$.

We have shown that every nilpotent subgroup of $G$ is abelian, and thus also cyclic of prime order, by Lemma 1. $\blacksquare$

---

Since every Sylow subgroup of $G$ is nilpotent, it is cyclic of prime order, which means $G$ has squarefree order.

Let $p$ be the smallest prime divisor of $|G|$. If $|G|$ has no other prime divisors, then $G\cong C_p$.

If $|G|$ does have other prime divisors, then we have by Burnside's normal $p$-complement theorem that $G$ has a normal $p$-complement, $N\triangleleft G$.

For every $x\in N$, $\mathbf{C}_G(x)=\langle x\rangle\subseteq N$. Thus, $N$ is a Frobenius kernel. Since Frobenius kernels are nilpotent, and nilpotent subgroups of $G$ are cyclic of prime order, it follows that $|N|=q$ for some prime $q\neq p$.

In this case, $|G|=pq$ with $p<q$. $G$ cannot be abelian, because abelian subgroups of $G$ are cyclic of prime order, so $G\cong C_q\rtimes C_p$, with $q\equiv 1\pmod p$.

Answer 1

There is a paper by Daniel Palacin ("Finite groups contain large centralizers", Israel Journal of Mathematics, 244,(2), (2021), 621-624) which proves (without the classification of finite simple groups, but making use of the odd order theorem), that if $G$ is any non-Abelian finite group, then there is a non-central element $x \in G$ such that $|G|< |C_{G}(x)|^{3}$. This may be viewed as being in the spirit of work begun by Brauer-Fowler. In terms of group orders, this result can't be expected to be improved much in its full generality, as is illustrated by the simple groups ${\rm SL}(2,2^{n}), n \geq 2$, for example.