Let $G$ be a finite group with the following property:
for each two of its subgroups $X,Y\subseteq G$ it holds either $X\cap Y=1$ or $X\subseteq Y$ or $Y\subseteq X$.
I want to show the following:
- If $H\leq G$ then either $|H|$ is a power of a prime or $|H|$ and $|G:H|$ are co-prime.
- If $1<N\trianglelefteq G$, then $G/N$ is nilpotent.
- If $N\trianglelefteq G$ and$N\neq G$, then $N$ is nilpotent.
$$$$
As for 2. :
So that $G/N$ is nilpotent we have to find a series of normal subgroups $$1\leq N_1\leq N_2\leq \dots \leq N_k=G/N$$ so that $N_{i+1}/N_i\subseteq Z((G/N)/N_i)$.
We have that $1<N\trianglelefteq G$.
Do we have to use the correspondence theorem to find he corresponding series for $G/N$ ?
But how can we show that quotient group belongs to the center?
But when we take the corresponding series, using the correspondence theorem, do we not get $1\leq G/N$ ?
That means that we have to check if $G/N\subseteq Z(G/N)$, or not?
Let's start with 1. Suppose that $|H|$ and $|G:H|$ are not coprime, so there is a prime $p$ dividing both. Let $P \in {\rm Syl}_p(H)$. Then there exists $Q \in {\rm Syl}_p(G)$ with $P < Q$, and the containment is strict because $p$ divides $|G:H|$. So $Q \cap H = P$, which is not equal to $Q$ and not equal to $1$. So $Q \cap H = H$ and hence $H=P$.