2-subnormal subgroup has a unique fastest ascending subnormal series

Statement

Statement with symbols

Suppose $H$ is a 2-subnormal subgroup (?) of a group $G$. Then, $H$ has a unique fastest ascending subnormal series; in other words, there is a subnormal series of $H$ of the form:

$H\triangleleft K\triangleleft G$

such that for any other subnormal series:

$H\triangleleft L\triangleleft G$

we have $L\leq K$.

Related facts

The analogous statement breaks down for 3-subnormal subgroups and groups of greater subnormal depth. In general, we do have a fastest descending subnormal series for any subnormal subgroup $H\leq G$ -- this is the series where each member is defined as the normal closure of $H$ in its predecessor.

Also, in the case that $N_{G}(H)$ is normal, we get $K=N_{G}(H)$. However, $N_{G}(H)$ is not necessarily a normal subgroup. In fact, it may even be an abnormal subgroup. Further information: 2-subnormal and abnormal normalizer not implies normal

Facts used

1. Normality is upper join-closed
2. Normality is strongly join-closed

Proof

Proof idea

The idea is to take $K$ as the join of all possible normal subgroups of $G$ in which $H$ is normal. We use facts (1) and (2) to show that $H\leq K\leq G$ forms a subnormal series for $H$. By definition, $K$ has been chosen as the largest possible subgroup for which we get a subnormal series.

Equivalently, $K$ is defined as the normal core in $G$ of the normalizer $N_{G}(H)$.

Proof details

Given: A group $G$, a 2-subnormal subgroup $H$ of $G$.

To prove: There exists a subgroup $K$ of $G$ such that $H\leq K\leq G$ is a subnormal series, and whenever $H\leq L\leq G$ is a subnormal series, $L\leq K$. Also, $K$ is the normal core of $N_{G}(H)$.

Proof: Define $K$ as the join of all subgroups $L$ having the property that $H\triangleleft L\triangleleft G$.

1. (Fact used: fact (1)): By fact (1), $H$ is normal in $K$.
2. (Fact used: fact (2)): By fact (2), $K$ is normal in $G$.
3. Thus, $H\leq K\leq G$ is a subnormal series. Further, by construction, we have that whenever $H\leq L\leq G$ is a subnormal series, $L\leq K$.

Note that since $H$ is normal in $K$, $K\leq N_{G}(H)$. Further, since $K$ is normal, $K$ is contained in the normal core of $N_{G}(H)$. It's also clear that $H$ is normal in the normal core of $N_{G}(H)$ and the normal core of $N_{G}(H)$ is normal in $G$, so maximality of $K$ yields that $K$ is the normal core of $N_{G}(H)$.