Given $S\unlhd\unlhd G$ (subnormal), where $S\unlhd H\leq G$. How can I construct a subnormal series from $S$ to $G$ that is $H$-invariant? Given $S\unlhd\unlhd G$ (subnormal), where $S\unlhd H\leq G$. How can I construct a subnormal series from $S$ to $G$ that is $H$-invariant (this is all the elements from the series are normalized by $H$). In a book they afirm that I can do this by taking normal closures (going to the distinguished subnormal series). But I'm not seing how.
Remark: All groups considered are finite.
 A: Let $G_0 = G$, and if $G_i$ is defined, define $G_{i+1}$ to be the $G_i$-normal closure of $S$. Since $S$ is subnormal, there is some $n$ such that $G_n=S$ [ let $S_n = S$ and $S_{i+1} \unlhd S_i$ and $S_0 = G$. Assume for induction that $G_i \leq S_i$. Then $S_{i+1}$ is normalized by $G_i$ and contains $S$, so $G_{i+1} \leq S_{i+1}$. Hence $G_i$ is the most swiftly descending subnormal series from $G$ to $S$. In particular, $S \leq G_n \leq S_n = S$. ]
$H$ normalizes $G_0$ by the assumption $H \leq G$ (which is overly strong). Suppose $H$ normalizes $G_i$. Then $H$ normalizes both $S$ and $G_i$, so it normalizes $G_{i+1}=\langle s^g : s \in S, g \in G_i \rangle$. Hence $H$ normalizes $G_i$ for all $i$.
[ Subnormal subgroups in infinite groups are much more complicated, but for this result, it doesn't matter whether the group is finite. Also, I believe there need not be a unique most swiftly ascending subnormal series, and I know the iterated normalizers do not always form a subnormal series. ]
