core of a subgroupCoreOfASubgroup2005-12-30 10:40:512007-06-13 14:21:21Definitioncore-freecorefreenormal-by-finitecore-free subgroupcorefree subgroupnormal-by-finite subgroup\usepackage{amssymb}
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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 \PMlinkname{quotient}{QuotientGroup} $G/\core{G}{H}$ 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).