Let $G$ be a finite group. Let the commutator of $x$ and $y$ is:
$$[x,y]= x^{-1}y^{-1}xy$$
$$[G, G] = \langle \{[x, y]\ \mid x,y\in G\}\rangle$$
$[G , G]$ is called the commutator subgroup of group $G$.
Now let us see the lower central series of group $G$.
$$G = L^{0}(G) \ge L^{1}(G) \cdots $$
Where $L^{i+1}(G) = [G,L^{i}(G)]$.
If the lower central series for $G$ terminates in $\{1\}$ then we say group is nilpotent.
Claim : A finite group $G$ is nilpotent iff if every maximal subgroup of $G$ is normal in $G$
Proof : Let me prove first that if every maximal subgroup of $G$ is normal in $G$ then group $G$ is nilpotent
Let $H$ be a maximal subgroup of $G$, by hypothesis it is normal so we get
$\{1\}\le H \le G$
As $H$ is maximal subgroup of $G$ it is not possible that we have some $H_1$ in which is non-trivial and a subgroup of $H$.
Is there anything more which I need to prove in this direction?
Other direction
If finite group $G$ is nilpotent then every subgroup of $G$ is maximal.
If $G$ is nilpotent then by above definition it admits an lower central series ..
I need a hint how to prove this direction
Best Answer
To be fair I don't see how $G \ge H \ge \{1\}$ helps you prove that $G$ is nilpotent.
First of all assume that $G$ isn't the trivial group, nor a $p$-group, as it's trivial that claim holds for those groups.
Anyway, here's a way how to prove the fact above. Assume that each maximal subgroup of $G$ is normal. Now let $P$ be a $p$-Sylow subgroup of $G$. Obviously $P \not = G$. Now we know that if each Sylow subgroup is normal in $G$, then it's nilpotent. So assume that $N[P] \not = G$. Then we have that $N[P] \le M < G$, where $M$ is some maximal subgroup of $G$. Now by Frattini's Argument we have:
$$G = MN_G[P] = M$$
which is a contradiction and hence $N[P] = G$ and so $P \lhd G$ and so $G$ is nilpotent.
For the other way first prove that nilpotent subgroups satisfy the nilpotent condition. Then let $M$ be any maximal subgroup of $G$, then we have that $M < N[M]$, but from the maximality of $M$ we must have $N[M] = G$ and hence $M \lhd G$ and hence the proof.