Full invariance does not satisfy intermediate subgroup condition

From Groupprops

This article gives the statement, and possibly proof, of a subgroup property (i.e., fully invariant subgroup) not satisfying a subgroup metaproperty (i.e., intermediate subgroup condition).
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Statement

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 $K$ .

Related facts

Intermediate subgroup condition

Other related facts

Full invariance does not satisfy image condition

Proof

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 $p t h$ powers in $G$ and $K = Ω 1 (G) = {(a, p b}$ 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 $p t h$ power under an endomorphism continues to be a $p t h$ 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

Further information: dihedral group:D8, subgroup structure of dihedral group:D8

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 $L$ .