Normal not implies image-potentially fully invariant

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 $$\rho(L) = H$$.

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 G 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 $$\varphi$$ is the retraction with kernel $$N$$ and image $$G_1$$, then $$\rho = \sigma \circ \varphi$$. 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 $$K$$.