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A comprehensive report on groups possessing a faithful irreducible character of small degree is contained in Walter Feit's report `The current situation in the theory of finite simple groups, Actes Co …

In post 6 is presented a weak result (it attained only for elementary abelian $P$). The Hall's result in post 3 is attained by many groups. It may be improved only in the case if we have an additional …

We claim that the class of $G$ is not bounded. Let $G$ be a $3$-group of maximal class and order $3^n>3^4$ containing no abelian subgroup of index $3$. Then $G$ has a metacyclic subgroup $M$ of index …

Addition to the above posts. If $n=2$, then our wreath product $G\cong\text{D}_8$ so it has exactly $3$ normal subgroups of index $2$. If $n>2$, then $(S_2)^n$ is the umique minimal normal subgroup of …

Some time ago I asked to classify the nonabelian $p$-groups of exponent $>p$ containing exactly $p$ maximal abelian subgroups of exponent $>p$. Below I'l prove that if $G$ is such a group, then $\exp( …

There is a number of papers dealing with groups with many normal subgroups. For example, let $G$ be a nonabelian group all of whose maximal abelian subgroups are normal. It is easy to see that then $ …

A s noted Prof. Robinson, this is false. However, this is true iff $L\not\le M\Phi(G)$. Indeed, let $\bar G=G/M\Phi(G)$; then $\bar L$ is a direct factor of $\bar G$ of order $p$. If $\bar G=\bar L\ti …

Let $G$ be an extraspecial group of order $p^{2n+1}$. Then $$ cp(G)=\frac{p^{2n}+p-1}{p^{2n+1}}=\frac1p+\frac{p-1}{p^{2n+1}}. $$ All maximal abelian subgroups have index $p^n$ in $G$. Hence, the answe …

Let $G$ be a noncyclic $p$-group. If $p>2$, then the number of cyclic subgroups of order $p^k>p$ in $G$ is a multiple of $p$ (G.A. Miller). If $p=2$ and, in addition, $G$ is not of maximal class, then …

If $G$ is an abelian $p$-group and $p^k\le\exp(G)$, then the number of cyclic subgropups of order $p^k$ in $G$ is $$ {\rm c}_k(G)=\frac{|\Omega_k(G)-\Omega_{k-1}(G)|}{(p-1)p^{k-1}}. $$ If the type of …

If for any prime $q$ dividing $\frac12(p^2+1)$ we have $q^3$ divides $\frac12(p^2+1)$, then the answer is `not'. This is purely number-theoretic question.

Let $H<G$ be maximal. If $H$ is abelian, then $G$ is solvable (Herstein). If $H$ is nilpotent of odd order, then $G$ is solvable (Thompson). If $H=P\times F$ is nilpotent and $P\in\text{Syl}_2(H)$ is …

Another proof of Strickland's result. Let $G$ be a group of order $p^n$ and $\nu(G)$ be the sum of orders of its cyclic subgroups. To prove that $\nu(G)=\sigma_1(G)$, we proceed by induction on $|G|$ …

It is easy to prove that if a nonabelian $p$-group $G$ of exponent $>p$ contains $<p$ maximal abelian subgroups of exponent $>p$, then the Hughes subgroup of $G$ is abelian of index $p$. PROBLEM. Cla …

In the case when the nonabelian $p$-group $G$ admits a partition by cyclic subgroups, an answer is known: either $\exp(G)=p$ or the Hughes subgroup of $G$ is a proper subgroup of $G$. In the general c …