PlanetMath: core of a subgroup

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core of a subgroup

(Definition)

Let be a subgroup of a group .

The core (or normal interior, or normal core) of in is the intersection of all conjugates of in :

$\displaystyle {{\mathrm{core}}}_{G}(H) = \bigcap_{x\in G}x^{-1}Hx.$

It is not hard to show that ${{\mathrm{core}}}_{G}(H)$ is the largest normal subgroup of contained in , that is, ${{\mathrm{core}}}_{G}(H)\trianglelefteq G$ and if $N\trianglelefteq G$ and $N\subseteq H$ then $N\subseteq{{\mathrm{core}}}_{G}(H)$ . For this reason, some authors denote the core by rather than ${{\mathrm{core}}}_{G}(H)$ , by analogy with the notation for the normal closure.

If ${{\mathrm{core}}}_{G}(H)=\{1\}$ , then is said to be core-free.

If ${{\mathrm{core}}}_{G}(H)$ is of finite index in , then is said to be normal-by-finite.

Let $\cal L$ be the set of left cosets of in . By considering the action of on $\cal L$ it can be shown that the quotient $G/{{\mathrm{core}}}_{G}(H)$ embeds in the symmetric group ${\mathrm{Sym}}({\cal L})$ . A consequence of this is that if is of finite index in , then ${{\mathrm{core}}}_{G}(H)$ is also of finite index in , and $[G:{{\mathrm{core}}}_{G}(H)]$ divides (the factorial of ). In particular, if a simple group has a proper subgroup of finite index , then must be of finite order dividing , 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).

"core of a subgroup" is owned by yark.

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