# Intermediately normal-to-characteristic implies intermediately subnormal-to-normal

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., intermediately normal-to-characteristic subgroup) must also satisfy the second subgroup property (i.e., intermediately subnormal-to-normal subgroup)
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This fact is an application of the following pivotal fact/result/idea: characteristic of normal implies normal
View other applications of characteristic of normal implies normal OR Read a survey article on applying characteristic of normal implies normal

## Statement

### Property-theoretic statement

The subgroup property of being intermediately normal-to-characteristic implies the subgroup property of being intermediately subnormal-to-normal.

## Definitions used

### Intermediately subnormal-to-normal subgroup

Further information: Intermediately subnormal-to-normal subgroup

A subgroup $H$ of a group $G$ is termed intermediately subnormal-to-normal if whenever $K$ is a subgroup containing $H$ such that $H$ is 2-subnormal in $K$, then $H$ is normal in $K$.

### Intermediately normal-to-characteristic subgroup

Further information: Intermediately normal-to-characteristic subgroup

A subgroup $H$ of a group $G$ is termed intermediately normal-to-characteristic if whenever $K$ is a subgroup containing $H$ such that $H$ is normal in $K$, $H$ is characteristic in $K$.

## Facts used

1. Characteristic of normal implies normal

## Proof

Given: A subgroup $H$ of a group $G$, such that $H$ is intermediately normal-to-characteristic in $G$.

To prove: For any subgroup $K$ of $G$ containing $H$, such that $H$ is 2-subnormal in $K$, $H$ is normal in $K$.

Proof: Since $H$ is 2-subnormal in $K$, there exists a subgroup $L$ such that $H \triangleleft L \triangleleft K$.

Now, $L$ is a subgroup of $G$, and $H$ is normal in $L$. Since $H$ is intermediately normal-to-characteristic in $G$, $H$ is characteristic in $L$.

Thus, $H$ is characteristic in $L$, that in turn is normal in $K$. By fact (1), $H$ is normal in $K$, completing the proof.