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

  • 1
    $\begingroup$ Downvoter care to explain $\endgroup$
    – 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.

  • 1
    $\begingroup$ Groups of order $p$ do have a maximal subgroup, the trivial subgroup. $\endgroup$
    – Derek Holt
    Feb 24 '18 at 11:27
  • $\begingroup$ @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 $G=P$. $\endgroup$
    – Stefan4024
    Feb 24 '18 at 11:30
  • $\begingroup$ @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 $G \ge H \ge {1}$. $\endgroup$
    – old
    Feb 24 '18 at 11:55
  • $\begingroup$ @old Yeah, the idea is right, but how do you prove that it's indeed a lower series? $\endgroup$
    – Stefan4024
    Feb 24 '18 at 12:03
  • $\begingroup$ @ 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 $G$. $\endgroup$
    – old
    Feb 24 '18 at 12:06

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