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I'm an undegraduate math student in Pontificia Universidad Católica de Chile.


1d
revised Non simplicity of a group of order $p^{100}q$ given some conditions.
added 70 characters in body
1d
comment Non simplicity of a group of order $p^{100}q$ given some conditions.
To use 2.- we must prove that there are 2 p sylow subgroups (If exists only one we are done) and for each pair of p Sylow subgroups i want to show that intersection is trivial. I dont believe that is true without abelianity.
1d
comment Non simplicity of a group of order $p^{100}q$ given some conditions.
I did the same in the exam. So probably 1.- is false. For 2.- I dont know how to do, my attempt is proving the hipotesys for 1 to find a normal q-subgroup
1d
comment Non simplicity of a group of order $p^{100}q$ given some conditions.
My teacher puts a $p$ in the exam, I also think it must be $q$. I think he is wrong. But in 2.- we probably need to find a $q$- Sylow using 1.-
1d
asked Non simplicity of a group of order $p^{100}q$ given some conditions.
Apr
14
accepted $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$
Apr
7
comment $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$
I understand now, I worked in details in your proof without difficulty. Ty.
Apr
7
comment $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$
Ty.................
Apr
6
comment $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$
Im almost sure that $x_i$ has maximal order in $G_{p_i}$. Is true that order of $g$ is the product of all $x_i$ orders? in that case I am right. For complements do you mean that $<g>=<x_1>\oplus \ldots \oplus <x_k>$?
Apr
6
comment $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$
I can follow your response unless "we can assume that there is a prime $p$..." What about $G=C_2\times C_{3^2}$? there are distinct primes here.
Apr
6
comment $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$
Why I can assume $p$-group?, I dont understand
Apr
6
comment $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$
Oh, teacher's proof uses it :(.
Apr
6
asked $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$
Mar
19
awarded  Popular Question
Jan
30
comment Solution Manual for Chapters 13 and 14, Dummit & Foote
No I don't since it is incomplete but I could upload most of chapter 13
Jan
25
awarded  Revival
Jan
25
comment Solution Manual for Chapters 13 and 14, Dummit & Foote
@JamesS.Cook I have done chapter 13 in spanish with my friends
Jan
19
comment Convergence = “closer and closer”
Or any positive strictly decreasing sequence which does not converges to $0$.
Jan
14
accepted Solution Manual for Chapters 13 and 14, Dummit & Foote
Jan
14
answered Solution Manual for Chapters 13 and 14, Dummit & Foote