Full invariance does not satisfy image condition

Statement
Suppose $$G$$ is a group, $$K$$ is a fully invariant subgroup of $$G$$, and $$\varphi:G \to H$$ is a surjective homomorphism. Then, $$\varphi(K)$$ need not be fully invariant in $$H$$.

Example of an Abelian group of prime-cube order
(This example uses additive notation).

Suppose $$G$$ is the direct product of a cyclic group $$A$$ of order $$p$$ and a cyclic group of order $$B$$ of order $$p^2$$. Define:


 * $$K = \Omega_1(G)$$ (see omega subgroups of a group of prime power order), i.e., $$K$$ is the subgroup comprising all the elements:

$$\{ x \in G \mid px = 0 \}$$.
 * $$\varphi$$ is the quotient map by the normal subgroup $$\mho^1(G)$$ (see agemo subgroups of a group of prime power order), i.e., $$\varphi$$ is the quotient map by the subgroup:

$$N := \{ y \in G \mid \exists x, px = y \}$$.
 * Observe that $$K$$ is fully invariant in $$G$$ (more generally, all omega subgroups are fully invariant). However, $$\varphi(K)$$ is a subgroup of order $$p$$ in $$\varphi(G)$$ which is elementary abelian of order $$p^2$$ -- hence $$\varphi(K)$$ is not fully invariant in $$\varphi(G)$$.

Example of a non-abelian group of prime-cube order
Let $$p$$ be an odd prime. Suppose $$A$$ is a cyclic group of order $$p^2$$ and $$B$$ is a cyclic group of order $$p$$, with $$B$$ acting on $$A$$ via multiplication by $$p+1$$. Then, the semidirect product of $$A$$ by $$B$$ is a non-Abelian group of order $$p^3$$. Call this group $$P$$. Define $$\Omega_1(P)$$ (see omega subgroups of a group of prime power order) as the subgroup generated by all elements of order $$p$$ in $$P$$. By the fact that Omega-1 of odd-order class two p-group has prime exponent, $$\Omega_1(P)$$ is a subgroup of prime exponent. This forces it to be a subgroup of order $$p^2$$ generated by the elements of $$B$$ and the multiples of $$p$$ in $$A$$. All the omega subgroups are fully characteristic, so $$\Omega_1(P)$$ is fully characteristic.

The center of $$P$$, namely $$Z(P)$$, simply comprises the multiples of $$p$$ in $$A$$. Thus, in the quotient map $$P \to P/Z(P)$$, the image of $$\Omega_1(P)$$ is cyclic of order $$p$$, while the whole group is elementary Abelian of order $$p^2$$. Thus:


 * $$\Omega_1(P)$$ is fully characteristic in $$P$$.
 * The image of $$\Omega_1(P)$$ in $$P/Z(P)$$ is not fully characteristic in $$P/Z(P)$$.