# About the two step centralizer

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.

• I think you have missed out some assumptions. $K_2 = G$ if and only if $\gamma_3(G)$ is trivial. Feb 18 at 21:02
• $K_2$ is trivial in any group of class at most $2$, so the asserted claim is false. Are you considering only nonabelian groups of maximal class, perhaps? Feb 18 at 21:03
• yes the paper consider the case of nonabelian p-groups of maximal class. I know that a p-group $G$ of order $p^n$ is of class at most $n-1$ but I can not prove the results Feb 18 at 21:11
• So then add that assumption to your post. Feb 18 at 21:39

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 $$K_2=G$$.

• Someone is downvoting my answers, one a day. Feb 19 at 4:43
• thank you so much for your help Feb 19 at 21:25