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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 ran …

(Addition to Isaacs' example) Let $G$ be a $p$-group and $d$ the minimal degree of representation of $G$ by permutations. Let $d<=p^n$. where $n$ is as small as possible. Then $G$ is a subgroup of $\S …

All the above mentioned results are well known (see \S 1 in `Groups of Prime Power Order', vol. 1 by Berkovich (there also given the autorship of the presented below results). Moreover, if a p-group G …

Let E be the elementary abelian group of order p^n. Then its Schur multiplier has rank n(n-1}/2 (Issai Schur). Therefore, the representation group of E (that group is special) does not satisfies ($*$) …

If a two-generator nonabelian p-group G satisfies the stated property, it is either of order p^3 or of class 2 with a cyclic subgroup of index p. To complete the classification, one can apply inductio …

Answer to the first question: an irregular group $G$ of order $p^{p+1}$ such that $|\Omega_1(G)|=p^{p-1}$ satisfies the condition. In particular, the minimal nonmetacyclic group of order $3^4$, and th …

(Essentially, Burnside) If $H$ is a $p$-group containing a nonabelian characteristic subgroup with cyclic center, then there is no $p$-group $G$ such that $H$ is a $G$-invariant subgroup of $\Phi(G)$ …

Answer on the question. Must be |Z(G)|\in{p,p^3} only. Asssume, however, that |Z(G)|=p^2. If A\le G is minimal nonabelian, then A\ne G (otherwise, |Z(G)|=p^3). In that case G=A*C_G(A), and it is easil …

Each finite $p$-group $H$ may be embedded in the Frattini subgroup of an appropriate $p$-group $G$ (for example, $G=H \wr C_p$; here $d(G)\le d(H)+1$. Then for this $G$, the Frattini subgroup $\Phi(G) …

Problem 1. Classify the nonabelian p-groups G such that Z(G) is contained in the union of minimal nonabelian subgroups of G. Problem 2 (old problem). Classify the p-groups covered by their minimal no …

In fact, if $|G|=p^n$ and $d(G)=d$, then $|\mathrm{Aut}(G)|$ divides the number $$(p^n-p^{n-d})(p^n-p^{n-d+1}) ... (p^n-p^{n-1}).$$ This result is due to P. Hall (1933) Proof. The above product is th …

The Sylow subgroup $\Sigma_n$ of the Symmetric group $S_{p^n}$ has only one maximal normal abelian subgroup, say $B$, for $p>2$ and $n>1$. As $\Sigma_n/B$ is isomorphic to $\Sigma_{n-1}$, the group $\ …