Normality satisfies intermediate subgroup condition

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 fact about::intermediate subgroup condition.

Related metaproperties satisfied by normality
Here are some related metaproperties that normality satisfies:

Related isomorphism theorems

 * Part of::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 $$G/H$$.
 * Part of::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

 * Weaker than::Left-inner implies intermediate subgroup condition
 * Weaker than::Left-extensibility-stable implies 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:

{| class="sortable" border="1" ! Property !! Meaning !! Proof that it dissatisfies intermediate subgroup condition !! Relation with normality (in meaning and proof)
 * Characteristic subgroup || invariant under all automorphisms || Characteristicity does not satisfy intermediate subgroup condition || The proof fails because automorphisms cannot always be extended to bigger groups (see extensible automorphisms problem)
 * Full invariance does not satisfy intermediate subgroup condition || invariant under all endomorphisms || Full invariance does not satisfy intermediate subgroup condition || The proof fails because endomorphisms cannot always be extended to bigger groups
 * ]
 * Full invariance does not satisfy intermediate subgroup condition || invariant under all endomorphisms || Full invariance does not satisfy intermediate subgroup condition || The proof fails because endomorphisms cannot always be extended to bigger groups
 * ]

Analogues in other algebraic structures

 * I-automorphism-invariance satisfies intermediate subalgebra 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.
 * Normality satisfies intermediate subloop condition: A normal subloop of an algebra loop is also normal in every intermediate subloop.

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
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, $$\sigma(x) = gxg^{-1}$$ 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 $$\sigma'$$ of the whole group $$G$$, whose restriction to $$K$$ is $$\sigma$$.
 * 4) Since $$H$$ is normal in $$G$$, $$\sigma'(H) \le H$$.
 * 5) Since the restriction of $$\sigma'$$ to $$K$$ is $$\sigma$$, and $$H \le K$$, we get $$\sigma(H) \le H$$.

Proof in terms of ideals
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.

Proof in terms of kernel of homomorphism
Given: A group $$G$$, a subgroup $$H$$ of $$G$$ that is the kernel of a homomorphism $$f:G \to L$$. A subgroup $$K$$ of $$G$$ containing $$H$$.

To prove: $$H$$ is the kernel of a homomorphism originating from $$K$$.

Proof: Let $$i:K \to G$$ be the inclusion map, and $$g = f \circ i$$. In other words, $$g:K \to L$$ is the restriction of $$f$$ to $$K$$. Then, $$g$$ is a homomorphism of groups (because it is a composite of two homomorphisms), and the kernel of $$g$$ is $$H$$, completing the proof.