Nonabelian subgroup of p-groups of maximal class

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.

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.