اشترك بالحزمة الذهبية واحصل على وصول غير محدود شمرا أكاديميا
تسجيل مستخدم جديدOn the subgroup structure of the hyperoctahedral group in six dimensions
132
0
0.0
(
0
)
اسأل ChatGPT حول البحث
ﻻ يوجد ملخص باللغة العربية
We investigate the subgroup structure of the hyperoctahedral group in six dimensions. In particular, we study the subgroups isomorphic to the icosahedral group. We classify the orthogonal crystallographic representations of the icosahedral group and analyse their intersections and subgroups, using results from graph theory and their spectra.
قيم البحث
اقرأ أيضاً
We study the class of finite groups $G$ satisfying $Phi (G/N)= Phi(G)N/N$ for all normal subgroups $N$ of $G$. As a consequence of our main results we extend and amplify a theorem of Doerk concerning this class from the soluble universe to all finite
groups and answer in the affirmative a long-standing question of Christensen whether the class of finite groups which possess complements for each of their normal subgroups is subnormally closed.
Recently Vaughan Jones showed that the R. Thompson group $F$ encodes in a natural way all knots, and a certain subgroup $vec F$ of $F$ encodes all oriented knots. We answer several questions of Jones about $vec F$. In particular we prove that the sub
group $vec F$ is generated by $x_0x_1, x_1x_2, x_2x_3$ (where $x_i, i=0,1,2,...$ are the standard generators of $F$) and is isomorphic to $F_3$, the analog of $F$ where all slopes are powers of $3$ and break points are $3$-adic rationals. We also show that $vec F$ coincides with its commensurator. Hence the linearization of the permutational representation of $F$ on $F/vec F$ is irreducible.
In this paper we describe the profinite completion of the free solvable group on m generators of solvability length r>1. Then, we show that for m=r=2, the free metabelian group on two generators does not have the Congruence Subgroup Property.
The congruence subgroup problem for a finitely generated group $Gamma$ asks whether the map $hat{Autleft(Gammaright)}to Aut(hat{Gamma})$ is injective, or more generally, what is its kernel $Cleft(Gammaright)$? Here $hat{X}$ denotes the profinite comp
letion of $X$. It is well known that for finitely generated free abelian groups $Cleft(mathbb{Z}^{n}right)=left{ 1right}$ for every $ngeq3$, but $Cleft(mathbb{Z}^{2}right)=hat{F}_{omega}$, where $hat{F}_{omega}$ is the free profinite group on countably many generators. Considering $Phi_{n}$, the free metabelian group on $n$ generators, it was also proven that $Cleft(Phi_{2}right)=hat{F}_{omega}$ and $Cleft(Phi_{3}right)supseteqhat{F}_{omega}$. In this paper we prove that $Cleft(Phi_{n}right)$ for $ngeq4$ is abelian. So, while the dichotomy in the abelian case is between $n=2$ and $ngeq3$, in the metabelian case it is between $n=2,3$ and $ngeq4$.
This contains a new version of the so-called non-commutative Gauss algorithm for polycyclic groups. Its results allow to read off the order and the index of a subgroup in an (possibly infinite) polycyclic group.
سجل دخول لتتمكن من نشر تعليقات
التعليقات
جاري جلب التعليقات
سجل دخول لتتمكن من متابعة معايير البحث التي قمت باختيارها
