Show that the groups are nilpotent

[mathmari]

Hey!

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:

1. If $H\leq G$ then either $|H|$ is a power of a prime or $|H|$ and $|G|$ are co-prime.
2. If $1<N\trianglelefteq G$, then $G/N$ is nilpotent.
3. 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 $G/N$ ? (Wondering)

But how can we show that quotient group belongs to the center? (Wondering)

But when we take the corresponding series, using the correspondence theorem, do we not get $1\leq G/N$ ? (Wondering)

That means that we have to check if $G/N\subseteq Z(G/N)$, or not? (Wondering)

 

[jvicens]

Hello!

I would like to provide some insight and guidance on your questions.

Firstly, let's look at the definition of a nilpotent group. A group $G$ is said to be nilpotent if it has a series of normal subgroups $1=N_0 \trianglelefteq N_1 \trianglelefteq \dots \trianglelefteq N_k = G$ such that $N_{i+1}/N_i\subseteq Z(G/N_i)$ for all $i=0,1,\dots,k-1$. This means that each quotient group $N_{i+1}/N_i$ is contained in the center of the quotient group $G/N_i$.

Now, let's consider your first question. If $H\leq G$, then either $|H|$ is a power of a prime or $|H|$ and $|G|$ are co-prime. To prove this, we can use the fact that $G$ has the property that for any two of its subgroups $X,Y\subseteq G$, either $X\cap Y=1$ or $X\subseteq Y$ or $Y\subseteq X$.

Suppose $|H|$ is not a power of a prime. This means that $|H|=ab$ for some positive integers $a$ and $b$ where $a,b>1$. Now, consider the subgroup $K=\langle x^a\rangle \leq H$, where $x\in H$ and $|x|=ab$. This means that $|K|=b$ and $K\subseteq H$.

Next, let's consider the quotient group $H/K$. By Lagrange's theorem, we know that $|H/K|=a$. Since $a$ and $b$ are co-prime, this means that $|H/K|$ and $|K|$ are co-prime as well.

Now, let's look at the second possibility. If $|H|$ is a power of a prime, then $|H|$ and $|G|$ are clearly co-prime. Thus, we have shown that either $|H|$ is a power of a prime or $|H|$ and $|G|$ are co-prime.

Moving on to your second question, if $1<N\trianglelefteq G$, then $G/N$