About the two step centralizer

Question

Let $G$ be a finite group. Define $\gamma_2(G)=[G,G]$ and $\gamma_{i+1}(G)=[\gamma_i(G),G]$ for all $i≥2$.

Let $G$ be a finite $p$-group of order $p^n$ and maximal class. For each $i$ with $2≤i≤n−2$, the $2$-step centralizer $K_i$ in $G$ is defined to be the centralizer in $G$ of $\gamma_i(G)/\gamma_{i+2}(G)$.

In some papers it is given without reference or proof that " $K_2$ coincides with $G$ if and only if $n = 3$".

I need a reference or proof of this results if there is any help.

Answer 1

For $p$-groups of maximal class, one reference is Leedham-Green and McKay's The Structure of Groups of Prime Power Order:

Proposition 3.1.4. Let $G$ be a finite $p$-group of order $p^n$ and of maximal class. Then the $2$-step centralizers $K_2,\ldots,K_{n-2}$ are maximal subgroups of $G$.

Proof. For $i$ satisfying $2\leq i\leq n-2$, there is an embedding of $G/K_i$ to the automorphism group of $\gamma_i(G)/\gamma_{i+2}(G)$ induced by conjugation. That quotient has order $p^2$ since $G$ is of maximal class. The automorphism group of a group of order $p^2$ has Sylow $p$-subgroups of order $p$, so $K_i$ has index $p$. $\Box$

That gives that the two-step centralizer $K_2$ is not $G$ if $n\geq 4$. If $n=3$, then $\gamma_2(G)$ is central and $\gamma_4(G)$ is trivial, so $K_2=G$.

If $G$ is not of maximal class then the "only if" clause fails, as clearly any group of class at most two will have $\gamma_2(G)$ central and $\gamma_4(G)$ trivial, so $K_2=G$.