Normality satisfies intermediate subgroup condition

This article gives the statement, and possibly proof, of a basic fact in group theory.
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This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) satisfying a subgroup metaproperty (i.e., intermediate subgroup condition)
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Statement

Verbal statement

If a subgroup is normal in the whole group, it is also normal in every intermediate subgroup of the group containing it.

Statement with symbols

Let $H\le K\le G$ be groups such that $H◃G$ (viz., $H$ is normal in $G$). Then, $H$ is normal in $K$.

Property-theoretic statement

The subgroup property of being normal satisfies the Intermediate subgroup condition (?).

Related facts

Related metaproperties satisfied by normality

Here are some stronger metaproperties that normality satisfies:

• Normality satisfies transfer condition: If $H$ is normal in $G$ and $K\le G$ is any subgroup, then $H\cap K$ is normal in $K$.
• Normality satisfies inverse image condition: If $\phi :K\to G$ is a homomorphism and $H$ is normal in $G$, ${\phi }^{-1}(H)$ is normal in $K$.

Here are some other related metaproperties that normality satisfies:

• Normality satisfies image condition
• Normality is upper join-closed

General conditions to ensure intermediate subgroup condition

• Left-inner implies intermediate subgroup condition
• Left-extensibility-stable implies intermediate subgroup condition

Intermediate subgroup condition for related properties

Here are some other properties that satisfy the intermediate subgroup condition:

• Central factor satisfies intermediate subgroup condition
• Direct factor satisfies intermediate subgroup condition
• Subnormality satisfies intermediate subgroup condition

Here are some that don't:

• Characteristicity does not satisfy intermediate subgroup condition
• Full characteristicity does not satisfy intermediate subgroup condition

Analogues in other algebraic structures

• I-automorphism-invariance satisfies intermediate subgroup condition: An I-automorphism in a variety of algebras is an automorphism expressible by a formula that is always guaranteed to yield automorphisms. In the variety of groups, the I-automorphisms are precisely the inner automorphisms.
• Ideal property satisfies intermediate subalgebra condition: In any variety of algebras, an ideal of an algebra is also an ideal in every intermediate subalgebra containing it.
• Ideal property satisfies intermediate subring condition in Lie rings: In a Lie ring, any ideal is also an ideal in every intermediate Lie subring.

Proof

Hands-on proof

Given: $H\le K\le G$ such that $H◃G$

To prove: $H◃K$: for any $g\in K$, $gH{g}^{-1}=H$.

Proof: Pick any $g\in K$. Since $K\le G$, $g\in G$. Further, since $H$ is normal in $G$ and $g\in G$, $gH{g}^{-1}=H$.

Proof in terms of inner automorphisms

This proof generalizes to the idea that I-automorphism-invariance satisfies intermediate subgroup condition over arbitrary varieties of algebras, and it also generalizes to results like left-inner implies intermediate subgroup condition and left-extensibility-stable implies intermediate subgroup condition.

The key idea here is that since inner automorphisms can be expressed by a formula that is guaranteed to yield an automorphism, any inner automorphism of a smaller subgroup extends to an inner automorphism of a bigger subgroup.

Given: $H\le K\le G$, such that $H$ is invariant under all inner automorphisms of $G$.

To prove: $H$ is invariant under all inner automorphisms of $K$.

Proof: Suppose $\sigma$ is an inner automorphism of $K$. Our goal is to show that $\sigma (H)\le H$.

1. Since $\sigma$ is inner in $K$, there exists $g\in K$ such that $\sigma =c_g$. In other words, $\sigma (x)=gx{g}^{-1}$ for all $x\in H$.
2. Since $K\le G$ and $g\in K$, we have $g\in G$.
3. The map $c_g:x\mapsto gx{g}^{-1}$ defines an inner automorphism ${\sigma }^{\prime }$ of the whole group $G$, whose restriction to $K$ is $\sigma$.
4. Since $H$ is normal in $G$, ${\sigma }^{\prime }(H)\le H$.
5. Since the restriction of ${\sigma }^{\prime }$ to $K$ is $\sigma$, and $H\le K$, we get $\sigma (H)\le H$.