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 $H\leq K\leq G$ be groups such that $H\triangleleft G$ (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 $K\leq G$ is any subgroup, then $H\cap K$ is normal in $K$.
• Normality satisfies inverse image condition: If $\varphi :K\to G$ is a homomorphism and $H$ is normal in $G$, $\varphi ^{-1}(H)$ is normal in $K$.

Here are some other related metaproperties that normality satisfies:

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

Related isomorphism theorems

• Fourth isomorphism theorem (also called the lattice isomorphism theorem or correspondence theorem): This states that if $H$ is normal in $G$, the quotient map $G\to G/H$ establishes a bijection between subgroups of $G$ containing $H$ (which is also a normal subgroup in each such subgroup) and subgroups of $G/H$.
• Third isomorphism theorem: This states that if $H\leq K\leq G$ and both $H,K$ are normal in $G$, then $H$ is normal in $K$, $K/H$ is normal in $G/H$, and $G/K\cong (G/H)/(K/H)$.

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: $H\leq K\leq G$ such that $H\triangleleft G$

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

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

Proof in terms of inner automorphisms

This proof method generalizes to the following results: I-automorphism-invariance satisfies intermediate subgroup condition over arbitrary varieties of algebras, 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: $H\leq K\leq G$, 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 $\sigma (H)\leq H$.

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

Proof in terms of ideals

This proof method generalizes to the following results: ideal property satisfies intermediate subalgebra condition over arbitrary varieties of algebras with zero.

The key idea here is to view the variety of groups as a variety with zero, i.e., a variety of algebras with a distinguished constant operation -- in this case, the identity element. The ideals in this variety are defined as follows: a subset $H$ of a group $G$ is an ideal if for any expression $\varphi (u_{1},u_{2},\dots ,u_{m},t_{1},t_{2},\dots ,t_{n})$ with the property that whenever all the $u_{i}$ are zero, the expression simplifies to zero, it is also true that whenever all the $u_{i}$ are in $H$ and the $t_{i}$s are in $G$, the expression yields a value in $G$.

It turns out that the ideals in the variety of groups are precisely the same as the normal subgroups (this is a consequence of the proof that the variety of groups is ideal-determined). We thus give the proof in terms of ideals in the variety of groups, assuming the equivalence.

Given: A group $G$, an ideal $H$ of $G$, a subgroup $K$ of $G$ containing $H$.

To prove: $H$ is an ideal of $K$. In other words, for any formula $\varphi (u_{1},u_{2},\dots ,u_{m},t_{1},t_{2},\dots ,t_{n})$ that simplies to the identity element whenever the $u_{i}$s are the identity element, we should have that the expression simplifies to a value inside $H$ whenever the $u_{i}$ are in $H$ and the $t_{i}$ are in $K$.

Proof: Notice that since the $t_{i}$ are in $K$, they are also in $G$. Since we know that $H$ is an ideal in $G$, we know by the property of $\varphi$ that $\varphi (u_{1},u_{2},\dots ,u_{m},t_{1},t_{2},\dots ,t_{n})\in H$. This completes the proof.