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

1. 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:

1. (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 $G_{i+1}\leq H_{n-i-1}\cap G_{i}\leq H_{n-i-1}$.
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 $G_{i}$.