Normality satisfies intermediate subgroup condition

This article gives the statement, and possibly proof, of a basic fact in group theory.
View a complete list of basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this

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)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about normal subgroup |Get facts that use property satisfaction of normal subgroup | Get facts that use property satisfaction of normal subgroup|Get more facts about intermediate subgroup condition

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

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 $H\le K\le G$ 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: $H\le K\le G$ 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 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: $H\le K\le G$, 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 $\sigma (H)\le H$.

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 $\phi (u_1,u_2,\dots ,u_m,t_1,t_2,\dots ,t_n)$ 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 $\phi (u_1,u_2,\dots ,u_m,t_1,t_2,\dots ,t_n)$ 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 $\phi$ that $\phi (u_1,u_2,\dots ,u_m,t_1,t_2,\dots ,t_n)\in H$. This completes the proof.