2-subnormal subgroup has a unique fastest ascending subnormal series

Statement with symbols
Suppose $$H$$ is a fact about::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 $$L \le K$$.

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 $$H \le G$$ -- 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.

Facts used

 * 1) uses::Normality is upper join-closed
 * 2) uses::Normality is strongly join-closed

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 $$H \le K \le G$$ 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 $$H \le K \le G$$ is a subnormal series, and whenever $$H \le L \le G$$ is a subnormal series, $$L \le K$$. 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, $$H \le K \le G$$ is a subnormal series. Further, by construction, we have that whenever $$H \le L \le G$$ is a subnormal series, $$L \le K$$.

Note that since $$H$$ is normal in $$K$$, $$K \le N_G(H)$$. 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 $$N_G(H)$$.