# Nilpotent iff every maximal subgroup is normal

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

• Downvoter care to explain – old Feb 24 '18 at 9:10

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.

• Groups of order $p$ do have a maximal subgroup, the trivial subgroup. – Derek Holt Feb 24 '18 at 11:27
• @DerekHolt Oh, yeah! I guess too often I overlook the whole group and the trivial subgroup. Anyway that case needs to be considered separately, as then in the case first part $G=P$. – Stefan4024 Feb 24 '18 at 11:30
• @Stefan4024 I thought to prove $G$ is nilpotent I have to just come with an lower central series that's why I come up with $G \ge H \ge {1}$. – old Feb 24 '18 at 11:55
• @old Yeah, the idea is right, but how do you prove that it's indeed a lower series? – Stefan4024 Feb 24 '18 at 12:03
• @ Stefan4024 I have taken $H$ as a maximal subgroup so it is normal in $G$ and second thing is $\{1\}$ Is subgroup of $H$ and $\{1\}$ is normal in $G$. – old Feb 24 '18 at 12:06