Normality satisfies intermediate subgroup condition

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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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Contents

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:

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

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, gHg − 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, 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 \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.

  1. Since σ is inner in K, there exists g \in K such that σ = cg. In other words, σ(x) = gxg − 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 gxg^{-1} defines an inner automorphism σ' of the whole group G, whose restriction to K is σ.
  4. Since H is normal in G, \sigma'(H) \le H.
  5. 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 ui are zero, the expression simplifies to zero, it is also true that whenever all the ui are in H and the tis 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 uis are the identity element, we should have that the expression simplifies to a value inside H whenever the ui are in H and the ti are in K.

Proof: Notice that since the ti 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.

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