# 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

## The definitions that we have to prove as equivalent

1. There exists an ascending chain $H = H_0 \le H_1 \dots H_n = G$ such that each $H_i$ is normal in $H_{i+1}$. The smallest possible $n$ for which such a chain exists is termed the subnormal depth of $H$.
2. Consider the descending chain $G_i$ defined as follows: $G_0 = G$ and $G_{i+1}$ is the normal closure of $H$ in $G_i$. Then, there exists an $n$ for which $G_n = H$. The smallest such $n$ is termed the subnormal depth of $H$.
3. Consider the sequence $K_i$ of subgroups of $G$ defined as follows: $K_0 = G$, and $K_{i+1} = [H,K_i]$ (the commutator), This sequence of subgroups eventually enters inside $H$. The number of steps taken is termed the subnormal depth of $H$.

## Facts used

1. Subnormal subgroup has a unique fastest descending subnormal series
2. Product with commutator equals join with conjugate: If $H \le G$ and $A$ is a subset of $G$, we have $\langle H, H^A \rangle = H[A,H]$.

## 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 $H$ of $G$. A descending subnormal series for $H$ in $G$ given by $G_0 = G$, and $G_{i+1}$ is the normal closure of $H$ in $G_i$. We have $G_n = H$.

To prove: Consider the sequence $K_0 = G$, $K_i = [H,G_i]$. Then, $K_n \le H$.

Proof: Observe that since $G_{i+1}$ is normal in $G_i$, we have:

$[G_{i+1}, G_i] \le G_{i+1}$.

Further, since $H \le G_{i+1}$, we get:

$[H,G_i] \le G_{i+1}$.

We now prove, by induction on $i$, that $K_i \le G_i$.

• The base case for induction is true, since for $i = 0$, $K_0 = G_0 = G$.
• The induction step: Suppose $K_i \le G_i$. Then $K_{i+1} = [H,K_i] \le [H,G_i] \le G_{i+1}$.

Thus, if $G_n = H$, we get $K_n \le H$.

### (3) implies (2)

Given: A subgroup $H$ of $G$. A descending chain of subgroups given by $K_0 = G$, and $K_{i+1} = [H,K_i]$. We have $K_n \le H$.

To prove: Define $G_0 = G$, and $G_{i+1}$ as the normal closure of $H$ in $G_i$. Then, $G_n = H$.

Proof: We do this by establishing a relation between $K_i$ and $G_i$ inductively, namely:

$G_i = HK_i$.

We prove this inductively. The base case is direct since both sides are equal to $G$ for $i = 0$. We thus need to prove the induction step, i.e., if $G_i = HK_i$, we need to prove that $G_{i+1} = HK_{i+1}$.

By fact (2), we have:

$H^{K_i} = H[K_i, H] = H[H,K_i] = HK_{i+1}$.

On the other hand, we have by induction that $G_i = HK_i$, so:

$G_{i+1} = H^{G_i} = H^{HK_i} = H^{K_i}$.

Putting these together, we get:

$G_{i+1} = HK_{i+1}$,

as desired.

Now that we have the general relation $G_i = HK_i$, it is clear that if $K_n \le H$, we have $G_n = H$.