Normal not implies potentially fully invariant

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) need not satisfy the second subgroup property (i.e., potentially fully invariant subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about potentially fully invariant subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not potentially fully invariant subgroup|View examples of subgroups satisfying property normal subgroup and potentially fully invariant subgroup

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. Equivalence of definitions of complete direct factor: This states that for a complete subgroup, being normal is equivalent to being a direct factor.
2. 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. Homomorph-containment satisfies intermediate subgroup condition

Proof

Example involving a complete group

Further information: Complete and potentially fully invariant implies homomorph-containing

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

Further information: Fully normalized and potentially fully invariant implies centralizer-annihilating endomorphism-invariant

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 $G$. Further information: D8 is not potentially fully invariant in D16