# 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 $H \le K \le G$ 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 related metaproperties that normality satisfies:

Metaproperty name Relation Proof of satisfaction Full statement
Transfer condition Stronger than intermediate subgroup condition Normality satisfies transfer condition If $H$ is normal in $G$ and $K \le G$ is any subgroup, then $H \cap K$ is normal in $K$
Inverse image condition 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 $K$
Image condition Injective maps replaced by surjective maps Normality satisfies image condition If $\varphi:G \to K$ is surjective and $H$ is normal in $G$, $\varphi(H)$ is normal in $K$
Upper join-closed subgroup property Normality is upper join-closed If $H \le G$ and $K_i, i \in I$ are all subgroups of $G$ in which $H$ is normal, $H$ is normal in the join of the $K_i$s

### 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 $G/K \cong (G/H)/(K/H)$.

### Intermediate subgroup condition for related properties

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

Property Meaning Proof that it satisfies intermediate subgroup condition Relation with normality (in meaning and proof)
Central factor every inner automorphism of whole group restricts to inner automorphism of subgroup Central factor satisfies intermediate subgroup condition both are examples of left-inner implies intermediate subgroup condition
Direct factor factor in an internal direct product Direct factor satisfies intermediate subgroup condition
Subnormal subgroup finite chain from subgroup to group, each normal in next Subnormality satisfies intermediate subgroup condition This actually follows from the stronger fact that normality satisfies transfer condition and transfer condition is composition-closed

Here are some that don't:

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 ]

## 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 subalgebra 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, $\sigma(x) = gxg^{-1}$ 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

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.

### 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.