363 reputation
110
bio website
location Université Kasdi Merbah Ouargla, Ouargla, Algeria
age 28
visits member for 1 year, 9 months
seen May 11 at 9:20

I'm a Ph.D candidate, I work on automorphism groups of finite p-groups.


Sep
24
awarded  Autobiographer
Jul
2
awarded  Curious
May
11
comment A question on $p$-central $p$-groups
@Geoff Robinson: Yes you are right. For $p=2$, one say that $G$ is $2$-central if every element of order dividing 4 is central. $A$ can be taken to be of exponent 4, so Alperin's result can be applied.
May
10
asked A question on direct limits of finite $p$-groups
May
10
asked A characterization of almost relatively free, finite $p$-groups
May
10
asked A question on $p$-central $p$-groups
May
6
comment Questions about some finite p-groups of coclass 2
Now I suppose that you mean $\Phi(G)=Z_{n-3}(G)$ in your second question. I think yes, $G/Z_{n-3}$ has order $p^2$ and it is not cyclic, thus it contains $\Phi(G)$. Since $G$ is 2-generated, it follows that $\Phi(G)=Z_{n-3}(G)$.
May
6
comment Questions about some finite p-groups of coclass 2
The conditions 4, 6, 7, can be deduced from 1 and 5. Also if $G$ is not regular then $cl(G) \geq p$. I suppose that $n$ is defined by $p^n=|G|$, if so then $G$ has class $n-2$; however if $\Phi(G)=Z_{n-4}$, then $Z_{n-3}(G)=G$ which means that $G$ has class $n-3$. I wish that you edit your question with this remarks in mind.
Apr
30
comment Is there a precise notion of “almost all” such that almost all finite groups are Galois groups of extensions of the rationals?
has an orbit in which all the members are $p$-nilpotent, then $|B_d|/|A_d| \rightarrow 1$ when $d \rightarrow \infty$. It is interesting to extend this definition to include all the primes.
Apr
30
comment Is there a precise notion of “almost all” such that almost all finite groups are Galois groups of extensions of the rationals?
For a finite $p$-group $P$, one can consider the class of all finite groups (up to isomorphism) containing $P$ as a Sylow $p$-subgroup; let us call this the orbit of $P$. It is shown by Henn and Priddy that almost all finite groups are $p$-nilpotent in the sense " for almost all $p$-groups $P$ of Frattini class $n$, any group in the orbit of $P$ is $p$-nilpotent". More precisely, If one denotes by $A_d$ the set of $p$-groups of Frattini class $n$that can be generated by $d$ elements, and one denotes by $B_d$ the subclass of $A_d$ having the property that any member $P$ of $B_d$,
Apr
19
answered Bound for the Frattini subgroup of a $p$-group
Apr
17
comment $2$-group with two isomorphic normal subgroups of index $4$ with non-isomorphic quotients
@Jeremy Rickard: you are right, thank you.
Apr
17
comment $2$-group with two isomorphic normal subgroups of index $4$ with non-isomorphic quotients
Ok, I think in a finite $p$-group $G$ such that $Aut(G)$ induces the full linear group on $G/\Phi(G)$, all the maximal subgroups of $G$ are isomorphic. Any relativley free finite $p$-group have the property above. I don't know if one can choose two maximal subgroups $N$ and $M$ such that the isomorphism between $N$ and $M$ (induced by an automorphism of $G$) takes a $G$ invariant maximal subgroup of $N$ whose quotient has exponent $4$ to a subgroup of $M$ whose quotient has exponent $2$. Another thing, any free Burnside group of exponent 4 is finite (thanks to Sanov).
Apr
16
comment $p$-groups with $\Omega_1(G)\leq\Phi(G)$
@Yves Cornulier : Ok you are confused with the notation. $\Omega_1(G)$ is always defined to be the subgroup generated by the elements of order $p$. The set of elements of order $p$ is usually denoted by the same notation , but one makes the index 1 between { }, and this agrees with Arturo's notation.
Apr
16
comment $p$-groups with $\Omega_1(G)\leq\Phi(G)$
@Hamid Mousavi: what kind of information you expect to have? I can say that for $p$ odd, the index of $G^p$ is determined by the order of the Frattini subgroup, and so the number of generators of $G$, these is not a very severe restriction. Also I think that any $p$-group is a qoutient of a group of the mentioned type.
Apr
16
comment $p$-groups with $\Omega_1(G)\leq\Phi(G)$
@Yves Cornulier: Don't be sorry, we are made to make mistakes. Also you may notice that $\Omega_1(G)$ does not have necessarily exponent $p$, so the reverse inclusion does not imply your claim so obviously.
Apr
16
comment $2$-group with two isomorphic normal subgroups of index $4$ with non-isomorphic quotients
I wonder if a 2-generated 2-group with all maximal subgroups isomorphic, and a non elementary abelian abelianized do the claim. Did you checked the free Burnside group on 2-generators and exponent 4.
Apr
16
comment $p$-groups with $\Omega_1(G)\leq\Phi(G)$
@Yves Cornulier : I don't understand your comment. It follows from it that such a group has exponent $p^2$ which I cannot see immediately.
Apr
13
comment Non split extension isomorphic (as a group) to a split extension
Perhaps this approach is useful. Assume that $B$ is a subgroup of $G$ which is isomorphic to $A$ as a $G$-module, such an isomorphism can be extended to an isomorphism $\phi : A \rightarrow B$ of $E$-modules. This induces an isomorphism between the abelian groups $Der(E,A)$ and $Der(E,B)$ in the obvious way. I wonder if $\phi$ induce a ring isomorphism or more generally if there is a ring isomorphism between $Der(E,A)$, $Der(E,B)$. If so Then $E$ splits over $B$. (The multiplication in $Der(G,A)$ is the composition of map)
Apr
13
comment Two $p$-groups whose automorphism groups have isomorphic Sylow $p$-subgroups
Thanks for your comments