# A certain centralizer group in finite group of order $3^5$ of maximal class contains every normal subgroup of index greater than $3$

Let $G$ be a finite group of order $3^5$ and maximal class, i.e. if $K_{i+1}(G) = [K_i(G), G]$ denotes the subgroups of the lower central series, we have $K_5(G) = 1$ and $K_4(G) = Z(G)$. Set $G_1 = C_G(K_2(G) / K_4(G))$, where $C_G(B/A) := \{ g \in G : hA = h^gA \mbox{ for all } h \in B \}$.

Then every normal subgroup of $G$ of index greater than $3$ is contained in $G_1$.

Why is this so? Do you have a proof of this fact?

If $G$ is a $p$-group of maximal class then, the only normal subgroups of $G$ are following

• $K_2(G)=[G,G], K_3(G)=[K_2(G),G], \cdots$ and they have index in $G$ respectively $p^2,p^3,\cdots$.

• The $p+1$ maximal subgroups containing $K_2(G)$.

Now if $|G|=3^5$, and if $H$ is normal subgroup of index $>3$, then it means, its index is $3^2$ or $3^3$ or $\cdots$, so it is inside $K_2(G)$ (as Derek Holt pointed out).

Now $K_2(G)/K_4(G)$ is of order $p^2$ and $G$ acts on it by conjugation, giving a homomorphism from $G$ to $Aut(K_2/K_4)$ and the automorphism group has order $p(p-1)$ or $(p^2-1)(p^2-p)$ according to $K_2/K_4$ is cyclic or of type $(p,p)$.

Then as $G$ is $p$-group, the image of homomorphism can have order $p$ or $1$, i.e. kernel (normal subgroup) has index $1$ or $p$; in any case, kernel contains a maximal subgroup, and hence it contains $[G,G]=K_2$.

In fact every normal subgroup of index greater than $3$ is contained in $K_2(G)$, which is contained in $G_1$.

That follows from the fact that nontrivial normal subgroups of $p$-groups intersect the centre nontrivially. Apply that to $G$, then to $G/K_4(G)$ and $G/K_3(G)$.

• What is $G_2$? I have given a definition of $G_1$, but I do not know $G_2$? Jan 8, 2016 at 20:12
• Guess $G_2 = C_G(K_3(G) / K_5(G))$ might be logical? Jan 8, 2016 at 20:19
• Sorry, I meant $K_2(G)$. Jan 8, 2016 at 21:12

Let $G$ be a $p$-group of maximal class. Then ${\rm Z}(G)$ is the unique minimal normal subgroup of $G$ therefore one may assume that $|G|>p^3$. Let $H$ be normal of index $>p$ in $G$. Then ${\rm Z}(G)\le H$. As $G/{\rm Z}(G)$ is of maximal class, we get, by induction, $H/{\rm Z}(G)\le (G/{\rm Z}(G))'=G'/{\rm Z}(G)$ so $H\le G'$. Since $G'<G_1$, we are done.