Full invariance does not satisfy intermediate subgroup condition

Statement with symbols
It is possible to have groups $$H \le K \le G$$ such that $$H$$ is a fully invariant subgroup of $$G$$ but $$H$$ is not a fully invariant subgroup of $$K$$.

Intermediate subgroup condition

 * Homomorph-containment satisfies intermediate subgroup condition
 * Homomorph-containing implies intermediately fully invariant
 * Characteristicity does not satisfy intermediate subgroup condition

Other related facts

 * Full invariance does not satisfy image condition

Abelian example of prime-cube order
Let $$p$$ be any prime. Consider the group:

$$G := \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}$$.

Let $$H = \mho^1(G) = \{ (0,pa) \}$$ be the set of $$p^{th}$$ powers in $$G$$ and $$K = \Omega_1(G) = \{(a,pb \}$$ be the set of elements of order $$1$$ or $$p$$. Then $$H \le K \le G$$, and:


 * $$H$$ is fully invariant in $$G$$: This is on account of it being an agemo subgroup -- the image of a $$p^{th}$$ power under an endomorphism continues to be a $$p^{th}$$ power.
 * $$H$$ is not fully invariant in $$K$$: In fact, $$K$$ is $$L \times H$$ where $$L = \{ (a,0) \}$$, so $$K$$ is an elementary abelian group of order $$p^2$$. In particular, $$K$$ has an automorphism interchanging the coordinates, and $$H$$ is not invariant under this automorphism.

Example of the dihedral group
Let $$G$$ be the dihedral group of order eight:

$$G := \langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle$$.

Let $$H = \langle a^2 \rangle$$ and $$K = \langle a^2, x \rangle$$. Then:


 * $$H$$ is fully invariant in $$G$$: In fact, $$H = \mho^1(G)$$, and also $$H$$ is the commutator subgroup of $$G$$, so $$H$$ is fully invariant in $$G$$.
 * $$H$$ is not fully invariant in $$K$$: In fact, $$K = H \times L$$ where $$L = \langle x \rangle$$, so $$K$$ is a Klein four-group and it admits an automorphism interchanging $$H$$ and $$L$$.