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Let denote by $V = V(c, s, t)$ the class of all finite $p$-groups such that $class(G) \leq c$, $expZ(G) \leq p^s$, and $G/Z(G)$ has a maximal normal abelian subgroup of rank $\leq t$. The class $V$ c …

Given a finite group $H$, How can one prove that the equation $Aut(X)=H$ has only finitely many solutions in the class of finite p-groups. (This would be the case if the divisibility conjecture is tru …

First I note that your assertion about Alperin's theorem is not true for $p=2$. I think the answer to your second question is negative. There is $p$-groups ($p>3$) having abelian subgroups of index …

Let $G$ be a finite $p$-group, $p$ odd, and let $A$ be a maximal elementary abelian normal subgroup of $G$. Assume that $x \in G$ centralizes $A$ and $x^p$ is a central element. Is it true that $x \in …

Given a free group $F$ on $d$ generators and a normal subgroup $H$ of $F$ whose index is finite of prime power order, is there a systematic way to find the numbers of generators of $H/[H,F]$ and of $H …

In a $p$-group $G$ of order $p^n$, any maximal abelian normal subgroup of order $p^m$ has index at most $p^{m(m-1)/2}$. To see this, observe that a maximal ablian normal subgroup of $G$ is self centr …

What can one say about a finite p-group $G$ in which the center of any maximal subgroup is equal to the center of $G$?. CH. J.Cossey and T.Hawkes, (Sets of p-powers as conjugacy class sizes, Proc. Ame …

For brevity, I denote by $F$ the free group on $n$ generators, $\lambda_k$ the $k$th terme of the $p$-lower series of $F$, and $N_k$ the relatively free group $F/\lambda_k$. Also I use the following …

How can one estimate the number of $p$-groups of order $\leq p^n$ that split over a normal abelian subgroup? Moreover, let $s(n,p)$ be the number of such groups, and let $f(n,p)$ denotes the number o …

Let be $G=F/F^p[F,F,F]$, with $F$ denotes the free group on $n$ generators. Then $G$ satisfies $Z(G) = \Phi (G)=G'$. If $x \in G-Z(G)$ then $C_G(x)= \langle x, Z(G) \rangle$ which is abelian as $C_ …

For your second question the answer is yes. The bound follows from Schreier's inequality: if $\Phi(G)$ has index $p^d$, then it follows that $\Phi(G)$ can be generated by $p^d(d-1)+1$ elements. Note …

Let $G$ be a finite $p$-group. Since we can embed $Z_2(G)/Z(G)$ in $Hom(G,Z(G))$, we have $d_2 \leq d(G)d(Z(G))$; where $d_2(G)=d(Z_2(G)/Z(G))$ and $d(G)$ denotes the minimal number of generators of $ …

Let $G$ be a group of odd order. It is known that if every central automophism of $G$ acts trivially on the center, then $G$ is purely non-abelain, this amounts to saying that every central endomorph …

Let $G$ be a group. Let us say that $G$ satisfies a generalized identity of degree $n$ if there exist $a_1,a_2,\dots a_n \in G$ such that $$x^{a_1}x^{a_2}\dots x^{a_n}=1,$$ for all $x\in G$. Assum …

Given a group $G$, we denote by $T(G)$ the subgroup generated by all (maximal) normal abelian subgroups of $G$. Let define the series $(T_i(G))$ by $T_0(G)=1$ and $T_{i+1}(G)/T_i(G)=T(G/T_i(G)$, and …