# Equivalence of definitions of subnormal subgroup

This article gives a proof/explanation of the equivalence of multiple definitions for the term subnormal subgroup

View a complete list of pages giving proofs of equivalence of definitions

## Contents

## The definitions that we have to prove as equivalent

- There exists an ascending chain such that each is normal in . The smallest possible for which such a chain exists is termed the subnormal depth of .
- Consider the descending chain defined as follows: and is the normal closure of in . Then, there exists an for which . The smallest such is termed the subnormal depth of .
- Consider the sequence of subgroups of defined as follows: , and (the commutator), This sequence of subgroups eventually enters inside . The number of steps taken is termed the
**subnormal depth**of .

## Facts used

- Subnormal subgroup has a unique fastest descending subnormal series
- Product with commutator equals join with conjugate: If and is a subset of , we have .

## Proof

Fact (1) covers the equivalence of definitions (1) and (2). We thus need to prove the equivalence of definitions (2) and (3).

### (2) implies (3)

**Given**: A subgroup of . A descending subnormal series for in given by , and is the normal closure of in . We have .

**To prove**: Consider the sequence , . Then, .

**Proof**: Observe that since is normal in , we have:

.

Further, since , we get:

.

We now prove, by induction on , that .

- The base case for induction is true, since for , .
- The induction step: Suppose . Then .

Thus, if , we get .

### (3) implies (2)

**Given**: A subgroup of . A descending chain of subgroups given by , and . We have .

**To prove**: Define , and as the normal closure of in . Then, .

**Proof**: We do this by establishing a relation between and inductively, namely:

.

We prove this inductively. The base case is direct since both sides are equal to for . We thus need to prove the induction step, i.e., if , we need to prove that .

By fact (2), we have:

.

On the other hand, we have by induction that , so:

.

Putting these together, we get:

,

as desired.

Now that we have the general relation , it is clear that if , we have .