315 reputation
18
bio website
location Université Kasdi Merbah Ouargla, Ouargla, Algeria
age 28
visits member for 1 year, 1 month
seen yesterday

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


1d
comment $2$-group with two isomorphic normal subgroups of index $4$ with non-isomorphic quotients
@Jeremy Rickard: you are right, thank you.
1d
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).
1d
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.
1d
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.
1d
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.
1d
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.
1d
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
Apr
7
revised Two $p$-groups whose automorphism groups have isomorphic Sylow $p$-subgroups
added 11 characters in body
Apr
7
comment Two $p$-groups whose automorphism groups have isomorphic Sylow $p$-subgroups
Ok, It is my mistake that I didn't noticed that $M$ is not abelian.
Apr
7
comment Two $p$-groups whose automorphism groups have isomorphic Sylow $p$-subgroups
@Jared: thanks for your comment; You may notice that $M$ is not abelian, so that $M \times E_n$ and $E_{n+3}$ cannot be isomorphic.
Apr
7
asked Two $p$-groups whose automorphism groups have isomorphic Sylow $p$-subgroups
Mar
2
awarded  Yearling
Feb
2
answered Index of agemo subgroups in $p$-groups
Feb
1
asked Index of agemo subgroups in $p$-groups
Nov
24
comment A finite $p$-group with certain properties
Absolutely. Actually I'm searching for a minimal counter example for math.stackexchange.com/questions/572308/…. Thinking about the same problem in successive nights is not healthy.
Nov
24
accepted A finite $p$-group with certain properties
Nov
24
revised A finite $p$-group with certain properties
added 67 characters in body
Nov
24
comment A finite $p$-group with certain properties
This is exactly the case that I want to avoid. The question is not very well formulated, I will edit it.