Let $p$ be a prime number. Let $H\le G$ be a subgroup of finite group $G$. We say $H$ is special if there exists a normal series of form:
$$
H=G_1 \unlhd G_2 \unlhd \dots \unlhd G_n=G
$$
where $G_i$ is a normal subgroup of $G_{i+1}$ and $[G_{i+1}:G_i]=p$ for each $i$. Prove that any subgroup of $G$ containing a special subgroup of $G$ is also special.
My idea is that we know the following facts:
(1) Let $K$ and $H$ be two special subgroups of $G$, then $H\cap K$ is also special.
(2) There is a special subgroup $L\le G$ contained in all special subgroups of $G$. And $L$ is a normal subgroup of $G$.
How to apply these two useful facts?
Best Answer
I think it can be done like this: By fact (2), we can fix a normal special group $H$ and denote $L$ the subgroup of $G$ containing $H$. Then under the map $\phi \colon G \to G/H$, $L$ is mapped to a subgroup $\phi(L)=L_1$ of $p$-group $G/H$.
If $L_1=\left\lbrace e \right\rbrace$ then $L=H$ and we are done. Otherwise, $L_1$ has cardinality $p^i$ for some $i >1$. By the First Sylow theorem, for each $n> j\ge i$, every subgroup of $G/H$ of order $p^j$ is normal in some subgroup of order $p^{j+1}$. Thus we have the sequence $$L_1 \unlhd L_2 \unlhd \ldots\unlhd L_n =G/H. $$ Then taking the inverse, we can find a sequence: $$L \unlhd \phi^{-1}(L_2) \unlhd \ldots\unlhd \phi^{-1}(L_n)=G. $$ The index $L_{i+1}/L_{i}$ is preserved and equal to $p$ due to the Third isomorphism theorem.