Subnormal subgroup has a unique fastest descending subnormal series

Statement
Suppose $$H$$ is a subnormal subgroup of $$G$$ of subnormal depth $$n$$; in other words, there exists a subnormal series for $$H$$ of the form:

$$H = H_0 \le H_1 \le \dots \le H_n = G$$,

where each $$H_i$$ is subnormal in $$H_{i+1}$$.

Then, consider the following descending chain of subgroups:


 * $$G_0 = G$$.
 * $$G_{i+1}$$ is the normal closure of $$H$$ in $$G_i$$.

Then $$G_n = H$$, and the series:

$$H = G_n \le G_{n-1} \le \dots \le G_1 \le G_0 = G$$

is a subnormal series for $$H$$: each $$G_{i+1}$$ is subnormal in $$G_i$$. Further, it is the fastest descending subnormal series for $$H$$ in $$G$$: we have $$G_i \le H_{n-i}$$ for all $$0 \le i \le n$$.

Related facts
This result shows that for any subnormal subgroup, there is a unique subnormal series for it that descends the fastest: the subnormal series defined by taking successive normal closures. It also shows that to compute the subnormal depth (i.e., the minimum possible length of a subnormal series) it suffices to compute the length of this particular subnormal series.

A subnormal subgroup need not in general have a fastest ascending subnormal series. Here are some related facts:


 * 2-subnormal subgroup has a unique fastest ascending subnormal series
 * 2-subnormal not implies hypernormalized: We cannot hope, in general, to construct a subnormal series for a subnormal subgroup by taking successive normalizers.
 * 2-subnormal and abnormal normalizer not implies normal: Even for a 2-subnormal subgroup, the normalizer need not be normal. In fact, there exist 2-subnormal subgroups that are not normal, but whose normalizer is abnormal.
 * 3-subnormal subgroup need not have a unique fastest ascending subnormal series
 * 2-subnormal and hypernormalized not implies 2-hypernormalized: Even if a subgroup is hypernormalized, the series of normalizers may not be the fastest ascending subnormal series.

Facts used

 * 1) uses::Normality satisfies transfer condition: If $$A, B \le C$$ and $$A$$ is normal in $$C$$, then $$A \cap B$$ is normal in $$B$$.

Proof
Given: $$H$$ is a subnormal subgroup of $$G$$ of subnormal depth $$n$$; in other words, there exists a subnormal series for $$H$$ of the form:

$$H = H_0 \le H_1 \le \dots \le H_n = G$$,

where each $$H_i$$ is subnormal in $$H_{i+1}$$.

To prove: Consider the following descending chain of subgroups:


 * $$G_0 = G$$.
 * $$G_{i+1}$$ is the normal closure of $$H$$ in $$G_i$$.

Then $$G_n = H$$, and the series:

$$H = G_n \le G_{n-1} \le \dots \le G_1 \le G_0 = G$$

is a subnormal series for $$H$$: each $$G_{i+1}$$ is subnormal in $$G_i$$. Further, it is the fastest descending subnormal series for $$H$$ in $$G$$: we have $$G_i \le H_{n-i}$$ for all $$0 \le i \le n$$.

Proof:


 * 1) (Facts used: fact (1)): $$G_i \le H_{n-i}$$: We prove this by induction on $$i$$. We know that $$G_0 \le H_n$$ by definition. Suppose we have $$G_i \le H_{n-i}$$. Then, $$H_{n-i-1}$$ is normal in $$H_{n-i}$$, so by fact (1), $$H_{n-i-1} \cap G_i$$ is a normal subgroup of $$H_{n-i}$$ and it contains $$H$$. By definition, $$G_{i+1}$$ is the smallest normal subgroup of $$G_i$$ in $$H$$, so we get $$G_{i+1} \le H_{n-i-1} \cap G_i \le H_{n-i-1}$$.
 * 2) $$G_n = H$$: This follows directly from step (1), setting $$i = n$$.
 * 3) The $$G_i$$s form a subnormal series: By definition, $$G_{i+1}$$ is the normal closure of $$H$$ in $$G_i$$, so $$G_{i+1}$$ is normal in $$G_i$$.