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Recall that a group of order $p^{m}$ is of maximal class, if $cl(G)=m-1>1$.

Let $G$ be a nonabelian p-group of maximal class wich have an abelian subgroup of index $p$. Note that:

  • $G$ has $p+1$ maximal subgroup. They are all of maximal class, exept one is not of maximal class noted by $G_{1}$ and called the fundamental subgroup of $G$ ($G_{1}=C_{G}(K_{2}(G)/K_{4}(G)) $).
    • every nonabelian subgroup of $G$ is of maximal class.

I found in exercice of the Book (p-groupe of maximal class $v1$ of Berkovich), that the number of nonabelian subgroups of index $p^n$ in $G$ equals $p^n$ provided $|G|\geq p^{n+3}$. why this is true?.

Any help would be appreciated so much. Thank you all.

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Note that your first bullet point is only true when $m \ge 4$, since otherwise all maximal subgroups of $G$ are abelian. If $m=3$ then there is nothing to prove so let's assume that $m \ge 4$.

Now the derived subgroup $[G,G]$ has index $p^2$ in $G$ and it is equal to the intersection of any two maximal subgroups of $G$. So $[G,G]$ is abelian. Hence any nonabelian subgroup of $G$ is contained in a unique maximal subgroup of $G$.

Now it's a straightforward induction on $m$. If $m=4$ then the result holds by your first bullet point. If $m>4$, then the $p$ maximal subgroups of $G$ are themselves of maximal class, and they have an abelian maximal subgroup (i.e. $[G,G]$), so they each have $p$ nonabelian maximal subgroups, which by the previous paragraph are all distinct. So we have a total of $p^2$ nonabelian subgroups of index $p^2$.

If $m>5$, then they each have $p$ nonabelian maximal subgroups giving a total of $p^3$ nonabelian subgroups of index $p^3$, and so on.

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