Normality satisfies intermediate subgroup condition

From Groupprops

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

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 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 $H \triangleleft G$ (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 $K \le G$ 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:

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

Intermediate subgroup condition for related properties

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

Here are some that don't:

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 $H \triangleleft G$

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

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

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

Since $σ$ is inner in $K$ , there exists $g \in K$ such that $σ = c g$ . In other words, $σ(x) = g x g - 1$ for all $x \in H$ .
Since $K \le G$ and $g \in K$ , we have $g \in G$ .
The map $c_g: x \mapsto gxg^{-1}$ defines an inner automorphism $σ'$ of the whole group $G$ , whose restriction to $K$ is $σ$ .
Since $H$ is normal in $G$ , $\sigma'(H) \le H$ .
Since the restriction of $σ'$ to $K$ is $σ$ , 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 $\varphi(u_1, u_2, \dots, u_m, t_1, t_2, \dots, t_n)$ 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.