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'core of a subgroup'
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| Title of object: |
core of a subgroup |
| Canonical Name: |
CoreOfASubgroup |
| Type: |
Definition |
| Created on: |
2005-12-30 10:40:51 |
| Modified on: |
2006-06-18 04:36:49 |
| Classification: |
msc:20A05 |
| Defines: |
core-free, corefree, normal-by-finite, core-free subgroup, corefree subgroup, normal-by-finite subgroup |
| Synonyms: |
core of a subgroup=core
core of a subgroup=normal core
core of a subgroup=normal interior |
Revision comment (for changes between this and next version):
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\newcommand{\C}{\mathbb{C}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newtheorem{theorem}{Theorem}
\def\genby#1{{\left\langle #1\right\rangle}}
\def\normal{\trianglelefteq}
\def\subgroup{\leq}
\DeclareMathOperator{\Sym}{Sym}
\DeclareMathOperator{\coreop}{core}
\def\core#1#2{{\coreop}_{#1}(#2)} |
Content:
\PMlinkescapeword{argument}
\PMlinkescapeword{divides}
\PMlinkescapeword{finite}
\PMlinkescapeword{property}
Let $H$ be a subgroup of a group $G$.
The \emph{core} (or \emph{normal interior}, or \emph{normal core}) of $H$ in $G$
is the intersection of all conjugates of $H$ in $G$:
\[ \core{G}{H} = \bigcap_{x\in G}x^{-1}Hx. \]
It is not hard to show that
$\core{G}{H}$ is the largest normal subgroup of $G$ contained in $H$,
that is, $\core{G}{H}\normal G$ and
if $N\normal G$ and $N\subseteq H$ then $N\subseteq\core{G}{H}$.
For this reason, some authors denote the core by $H_G$
rather than $\core{G}{H}$,
by analogy with the notation $H^G$ for the normal closure.
If $\core{G}{H}=\{1\}$,
then $H$ is said to be \emph{core-free}.
If $\core{G}{H}$ is of finite index in $H$,
then $H$ is said to be \emph{normal-by-finite}.
Let $\cal L$ be the set of left cosets of $H$ in $G$.
By considering the action of $G$ on $\cal L$ it can be shown that
the quotient $G/\core{G}{H}$ embeds in the symmetric group $\Sym({\cal L})$.
A consequence of this is that if $H$ is of finite index in $G$,
then $\core{G}{H}$ is also of finite index in $G$,
and $[G:\core{G}{H}]$ divides $[G:H]!$ (the factorial of $[G:H]$).
In particular, if a simple group $S$ has a proper subgroup of finite index $n$,
then $S$ must be of finite order dividing $n!$,
as the core of the subgroup is trivial.
It also follows that
a group is virtually abelian if and only if it is abelian-by-finite,
because the core of an abelian subgroup of finite index
is a normal abelian subgroup of finite index
(and the same argument applies if `abelian' is replaced by
any other property that is inherited by subgroups). |
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