Let G be a p-group of maximal class and order $p^n\ne2^3$. Then rank of G is at most p. In that case, $L(G)=\Phi(G)$ has order $p^{n-2}$. As $n$ is unbouded, the answer on the question is not. The rank of $G$ is at most $p$ (See Y. Berkovich, Groups of Prime Power Order, 1, Exercise 9.13. If rank is $p>2$, then $G$ is isomorphic to a Sylow $p$-subgroup $\Sigma_{p^2}$ of the symmetric group of degree $p^2$.

Let G be a p-group of maximal class and order $p^n\ne2^3$. Then rank of G is at most p. In that case, $L(G)=\Phi(G)$ has order $p^{n-2}$. As $n$ is unbouded, the answer on the question is not. The rank of $G$ is at most $p$ (see Y. Berkovich, Groups of Prime Power Order, 1, Exercise 9.13). If the rank is $p>2$, then $G$ is isomorphic to a Sylow $p$-subgroup $\Sigma_{p^2}$ of the symmetric group of degree $p^2$. For $p=2$ this is not true.

Let G be a p-group of maximal class and order $p^n\ne2^3$. Then rank of G is at most p. In that case, $L(G)=\Phi(G)$ has order $p^{n-2}$. As $n$ is unbouded, the answer on the question is not.

Let G be a p-group of maximal class and order $p^n\ne2^3$. Then rank of G is at most p. In that case, $L(G)=\Phi(G)$ has order $p^{n-2}$. As $n$ is unbouded, the answer on the question is not. The rank of $G$ is at most $p$ (See Y. Berkovich, Groups of Prime Power Order, 1, Exercise 9.13. If rank is $p>2$, then $G$ is isomorphic to a Sylow $p$-subgroup $\Sigma_{p^2}$ of the symmetric group of degree $p^2$.