Normal not implies image-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., image-potentially fully invariant subgroup)
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Statement

It is possible to have a normal subgroup $H$ of a group $G$ that is not an image-potentially fully invariant subgroup of $G$, i.e., it is not possible to have a surjective homomorphism $\rho :K\to G$ and a subgroup $L$ of $K$ such that $\rho (L)=H$.

Related facts

Similar facts

• Normal not implies potentially fully invariant
• Normal not implies potentially verbal

Opposite facts

• NIPC theorem: This states that every normal subgroup is image-potentially a characteristic subgroup.
• NPC theorem: This states that every normal subgroup is potentially a characteristic subgroup.

Proof

Suppose $G$ is the free group of rank two and $H$ is a normal subgroup of $G$ that is not a fully invariant subgroup of $G$. In other words, there exists an endomorphism $\alpha$ of <mah>G</math> such that $\alpha (H)$ is not contained inside $H$.

Suppose $\rho :K\to G$ is a surjective homomorphism and $L$ is a subgroup of $K$ such that $\rho (L)=H$. We show that $L$ is not fully invariant in $K$.

Suppose $N$ is the kernel of $\rho$. Since $G$ is a free group, $N$ is a complemented normal subgroup, so there exists a complement $G_1$ to $N$ in $K$ with an isomorphism $\sigma :G_1\to G$ such that if $\phi$ is the retraction with kernel $N$ and image $G_1$, then $\rho =\sigma \circ \phi$. In particular, restricted to $G_1$, $\rho =\sigma$.

Now, consider the endomorphism $\beta$ of $K$ defined as $\beta ={\sigma }^{-1}\circ \alpha \circ \rho$. Then, we see that $\beta (L)={\sigma }^{-1}(\alpha (\rho (L))={\sigma }^{-1}(\alpha (H))$. Thus, $\rho (\beta (L))=\sigma (\beta (L))=\alpha (H)$. But $\alpha (H)$ is not contained in $H$, so $\rho (\beta (L))$ is not contained in $H$, so $\beta (L)$ is not contained in ${\rho }^{-1}(H)$. In particular, $\beta (L)$ is not contained in $L$. Hence, $L$ is not fully invariant in $K$.