Full invariance does not satisfy intermediate subgroup condition

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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 $K$.

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 $p^{th}$ 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

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 $L$.