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 $H=H_{0}\leq H_{1}\dots H_{n}=G$ 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 $H\leq G$ 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, $K_{n}\leq H$.

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

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

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

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

We now prove, by induction on $i$, that $K_{i}\leq G_{i}$.

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

Thus, if $G_{n}=H$, we get $K_{n}\leq H$.

(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 $K_{n}\leq H$.

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 $K_{n}\leq H$, we have $G_{n}=H$.