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:

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

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:

$H = G_{n} \leq G_{n - 1} \leq \dots \leq G_{1} \leq G_{0} = G$

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 $G_{i} \leq H_{n - i}$ for all $0 \leq i \leq n$ .

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

Normality satisfies transfer condition: If $A, B \leq C$ 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:

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

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:

$H = G_{n} \leq G_{n - 1} \leq \dots \leq G_{1} \leq G_{0} = G$

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 $G_{i} \leq H_{n - i}$ for all $0 \leq i \leq n$ .

Proof:

(Facts used: fact (1)): $G_{i} \leq H_{n - i}$ : We prove this by induction on $i$ . We know that $G_{0} \leq H_{n}$ by definition. Suppose we have $G_{i} \leq H_{n - i}$ . 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 $G_{i + 1} \leq H_{n - i - 1} \cap G_{i} \leq H_{n - i - 1}$ .
$G_{n} = H$ : This follows directly from step (1), setting $i = n$ .
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 $G_{i}$ .