Fully normalized subgroup

Symbol-free definition
A subgroup of a group is termed fully normalized if it satisfies the following equivalent conditions:


 * Every automorphism of the subgroup lifts to an inner automorphism of the whole group
 * The Weyl group (viz the image of the canonical map from the normalizer of the subgroup to its automorphism group), is the whole automorphism group

Definition with symbols
A subgroup $$H$$ of a group $$G$$ is termed fully normalized if it satsifies the following equivalent conditions:


 * For any automorphism $$\sigma$$ of $$H$$, there is an element $$g$$ in $$G$$ such that $$\sigma(x) = gxg^{-1}$$ for all $$g$$ in $$G$$.
 * The map $$c: N_G(H) \to Aut(H)$$ that sends $$g \in N_G(H)$$ to the automorphism $$h \mapsto ghg^{-1}$$ is surjective (that is, its image is the whole of $$Aut(H)$$).

Conjunction with other properties

 * Normal fully normalized subgroup: This can be viewed as a subgroup where the map $$c:G \to Aut(H)$$ given by conjugation is well-defined and surjective.

Trimness
The subgroup property of being fully normalized is trivially true, that is, the trivial subgroup is fully normalized in any group. It is not identity-true. In fact, a group satisfies the property as a subgroup of itself if and only if every automorphism of the group is inner.

Left-realized
Every group can be embedded as a fully normalized subgroup of some group (in fact, as a normal fuly normalized subgroup of itself). A natural example of such a group is the holomorph of the given group, which is the semidirect product of the group with its automorphism group. We say that the property of being fully normalized is a left-realized subgroup property.

If $$H$$ is fully normalized in $$K$$, and $$G$$ is a group containing $$K$$, then $$H$$ is fully normalized in $$G$$. Thus, the property of being fully normalized is a right-hereditary subgroup property.

Since the property of being fully normalized is right-hereditary, it is automatically transitive.