# Let $G$ be a $2$-group and suppose the centralizer of some element of order two has order at most four, then $G$ has maximal class

Let $G$ be a $2$-group of order $|G| \ge 4$ and $H \le G$ be a subgroup of order $2$, i.e. $|H| = 2$. Suppose we have $|C_G(H)| \le 4$. Then $G$ has maximal class.

Do you know a proof of this fact? I saw it used in an article, but without giving any further references.

Denote by $\gamma_1(G) := G, \gamma_{i+1} := [\gamma_i(G), G]$ the lower central series. A group is called nilpotent if $\gamma_{k+1}(G) = 1$ for some $k$, the smallest such $k$ is called the nilpotency class, or just class, of $G$. If $G$ is nilpotent of class $k$ with $|G| = p^{k+1}$ then we have $|\gamma_1(G) : \gamma_2(G)| = p^2$ and $|\gamma_i(G) : \gamma_{i+1}(G)| = p$ for all $i = 2,\ldots, k$, then we say that $G$ has maximal class. Another series is $Z_0(G) := 1$ and $Z_{i+1}(G) / Z_i(G) = Z(G/Z_i(G))$, the upper central series, then if $G$ is nilpotent of class $k$ we have $$Z_k(G) = G \mbox{ and } Z_{k-1}(G) \ne G$$ But I do not see how these definitions are connected to the centralizer of some order two subgroup?

We suppose that $H\subset Z(G)$, then the cardinal of $G$ is inferior to $4$. The problem is trivial.
Suppose that $H$ is not contained in $Z(G)$, then $C_G(H)$ contains $Z(G)$ and $H$, this implies that the cardinal of $Z(G)$ is 2, otherwise the cardinal of $C_G(H)\geq 8$.
Let $G'=G/Z(G)$ and $H'$ the projection of $H$ to $G'$ by $p:G\rightarrow G/Z(G)$. We are going to show that the cardinal of $C_{G'}(H')\leq 4$. Let $x',y'\in C_{G'}(H')$, consider $x,y\in G$ such that $p(x)=x'$ and $p(y)=y'$. Let $h\in H$, write $h'=p(h)$. We have $[x',h']=1$. It implies $[x,y]\in Z(G)$. If $[x,h]=1$, then $x\in C_G(H)$, if $[x,h]\neq 1$, suppose also that $[y,h]\neq 1$, then since the cardinal of $Z(G)$ is $2$, $[x,h]=[y,h]$ and $[xy^{-1},h]=1$. It implies that $xy^{-1}\in C_G(H)$. This shows that the elements of $p^{-1}(C_{G'}(H')$ are elements of $C_G(H)$ + elements $xu$ where $x$ is a fixed element such that $[x,h]$ is the non trivial element of $Z(G)$ and $u\in C_G(H)$. We deduce that the cardinal of $p^{-1}(C_{G'}(H'))\leq 8$ since $Z(G)\subset p^{-1}(C_{G'}(H'))$, we deduce that the cardinal of $C_{G'}(H')\leq 4$. We can thus conclude with a recursive argument.
• I do not understand your reasoning... If $p^{-1}(C_G'(H')) = C_G(H) \cup C_G(H)u$ with $Z(G) = \{1,u\}$ then as $Z(G) \le C_G(H)$ we have $p^{-1}(C_G'(H')) = C_G(H)$ and this would even imply $|C_{G'}(H')| \le 2$ and so $Z(G') = C_{G'}(H') = H'$ which gives $|G'| \le 4$ as you wrote in your first sentence. Also why you could suppose $[y,h] \ne 1$, what if $[y,h] = 1$? – StefanH Feb 14 '16 at 14:12