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6
votes
Accepted
Finite $p$-groups of maximal class whose generators have order $p$
The classification of such groups is as difficult as the classification of all $p$-groups of maximal class. Note that, for the latter problem, beside the cases where $p=2,3$ that were settled by Blac …
5
votes
3
answers
572
views
Normal abelian subgroups in p-groups
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 …
4
votes
Accepted
p-group with abelian centralizer
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_ …
4
votes
1
answer
419
views
Generators of p-groups
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 $ …
4
votes
1
answer
447
views
Number of generators of the automorphism group of an abelian group
Let $G$ be a finite abelian $p$-group. What is known about the minimal number of generators of a $p$-sylow of $Aut(G)$? is it bounded in terms of $d(G)$ the minimal number of generators of $G$ (and pe …
4
votes
2
answers
430
views
Index of agemo subgroups in $p$-groups
Having a finite $p$-group $G$ ($p$ odd). we denote by $\Omega_1(G)$ the subgroup generated by all the elements of $G$ of order dividing $p$.
Is there an example of such a group $G$, such that $|G: …
4
votes
2
answers
646
views
A question on $p$-central $p$-groups
Let $p$ be a fixed prime. A group $G$ is termed $p$-central if every element of order $p$ in $G$ lies in the center.
Having a finite $p$-group $G$ of rank $k$ (the least integer, such that every sub …
3
votes
Index of agemo subgroups in $p$-groups
This answer is based on Holt's counter example.
In a $p$-group of maximal class $G$, it is known that:
$|G:G^p|=p^p$ and $\Omega_1(G)$ has either order $p^{p-1}$ or index $p$.
Now we take $G$ of ma …
2
votes
Torsion in profinite groups
Too long for a comment.
I note first that I made an attempt to reduce the problem to the case where $K$ is normal, but it turned out to be false; I'm thankful to Ian Agol for his discussion. The cas …
2
votes
0
answers
129
views
Non left $k$-Engel elements in a nilpoent group always generate this group
Given a finite nilpotent group $G$ and let us denote by $L_n(G)$ the set of left $n$-Engel elements in $G$.
Assume that $n$ is the smallest positive integer such that $L_n(G)=G$.
Is it true that $G$ …
2
votes
0
answers
201
views
Two $p$-groups whose automorphism groups have isomorphic Sylow $p$-subgroups
Fix a prime $p$, and let $M$ be the unique nonabelian group of order $p^3$ and exponent $p$. Let us denote by $E_n$ the elementary abelian group of rank $n$.
Is it true that $\operatorname{Aut}(M …
2
votes
1
answer
105
views
A characterization of almost relatively free, finite $p$-groups
Let $G$ be a finite minimally $d$-generated $p$-group.
If $G$ is relatively free, that is $G$ is a quotient of the free group $F$ on $d$ generators by a fully invariant subgroup of $F$, then the orde …
1
vote
0
answers
133
views
The number of $p$-groups of order $\leq p^n$ that split over a normal abelian subgroup
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 …
1
vote
1
answer
306
views
A finite $p$-group with certain properties
Is there a finite $p$-group $G$ such that :
(a) $G= \langle A,x,y \rangle$, with $G/Z(G)$ has exponent $p$, $A$ is a maximal abelian normal subgroup of $G$, and $G/A$ has order $p^2$ (thus it is elem …
1
vote
The rank of indecomposable finite abelian 2-group
No, the rank of $G$ could be arbitrary large. Here is a counter-example that works for all primes $p$. Fix a positive integer $n$, and let $G$ be the direct sum of $n$ copies of $\mathbb{Z}/p^2\mathb …