Equivalence of definitions of subnormal subgroup

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) uses::Subnormal subgroup has a unique fastest descending subnormal series
 * 2) uses::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 $$G_n = H$$.