Normal not implies potentially fully invariant

Statement
It is possible to have a normal subgroup $$H$$ of a group $$G$$ that is not a potentially fully invariant subgroup of $$G$$ -- in other words, there is no group $$K$$ containing $$G$$ such that $$H$$ is a fully invariant subgroup of $$K$$.

Related facts

 * Normal not implies image-potentially fully invariant
 * Normal not implies potentially verbal
 * NPC theorem: Normal equals potentially characteristic.
 * Normal equals potentially normal-subhomomorph-containing
 * Complete and potentially fully invariant implies homomorph-containing
 * Fully normalized and potentially fully invariant implies centralizer-annihilating endomorphism-invariant

Facts used

 * 1) uses::Equivalence of definitions of complete direct factor: This states that for a complete subgroup, being normal is equivalent to being a direct factor.
 * 2) uses::Equivalence of definitions of fully invariant direct factor: This states that for a direct factor, being a fully invariant subgroup is equivalent to being a homomorph-containing subgroup.
 * 3) uses::Homomorph-containment satisfies intermediate subgroup condition

Example involving a complete group
Let $$A$$ be a nontrivial complete group. Define $$G := A \times A$$ and $$H := A \times \{ e \}$$. Clearly, $$H$$ is a normal subgroup of $$G$$.

Suppose $$K$$ is a group containing $$G$$, such that $$H$$ is fully invariant in $$K$$. In particular, $$H$$ is normal in $$K$$. Since $$H$$ is complete, it is a direct factor, so there exists a group $$C$$ that is a complement to $$H$$, so $$K = H \times C$$ as an internal direct product. Further, since $$G/H \cong H \cong A$$ is a subgroup of $$K/H$$, $$C$$ has a subgroup, say $$B$$, isomorphic to $$A \cong H$$.

Then, consider the endomorphism $$\alpha$$ of $$K$$ that sends $$C$$ to the trivial subgroup and $$H$$ isomorphically to the subgroup $$B$$. This endomorphism does not send $$H$$ to within itself.

More general example
More generally, suppose $$H$$ is a fully normalized subgroup of $$G$$ that is normal in $$G$$, but such that there is a homomorphism $$\theta: G \to G$$ whose kernel contains $$C_G(H)$$ such that $$\theta(H)$$ is not contained in $$H$$ (in other words, $$H$$ is not a centralizer-annihilating endomorphism-invariant subgroup). Then, $$H$$ is not a potentially fully invariant subgroup of $$G$$.

Examples include:


 * $$G$$ is the dihedral group of order 16, say $$G := \langle a,x \mid a^8 = x^2 = e, xax = x^{-1}\rangle$$, and $$H = \langle a^2,x \rangle$$. Then, $$C_G(H) = \langle a^4 \rangle$$ and we have a homomorphism $$\theta$$ from $$G$$ to $$G$$ such that $$\theta(a) = a^2, \theta(x) = ax$$ with kernel containing $$C_G(H)$$ and with $$\theta(H)$$ not contained in $$H$$. Thus, $$H$$ is not a potentially fully invariant subgroup of $$G$$.