Every proper maximal subgroup of a $p$-group $P$ is normal and has index $p$.

Question

Every proper maximal subgroup of a $p$-group $P$ is normal and has index $p$.

I tried to search online by I can't get a complete proof.

Take $M$ to be maximal and $Z$ to be central subgroup of order $p$.

Case 1: $Z\subset M$
We prove by induction on $|P|$.
For the base case, $|P|=p$.
There is no maximal subgroup in $P$, hence the result holds

Assume that the result holds for all $p$-group $Q$ with $|Q|<|P|$
Note that $|Z|>1$. Hence $|P/Z|<|P|$.
Suppose that $M/Z$ is not a maximal subgroup of $P/Z$.
Then there exists a subgroup $H$ such that $M/Z\subset H/Z \subset P/Z$.
But this also implies that $M\subset H \subset P$.
This means that $M$ is not maximal, a contradiction.
Hence $M/Z$ is a maximal subgroup of $P/Z$.
By induction hypothesis, $M/Z$ has index $p$ and $M/Z\trianglelefteq P/Z$.
First, we can conclude that $M \trianglelefteq P$.
Second, note that $[P/Z:M/Z]=p$. Hence $[P:M]=p$.

Case 2: $Z\not\subset M$
This implies that $P=ZM$.
Let $p=zm\in P$ where $z\in Z$ and $m\in M$.
Clearly, $p^{-1}Mp=M$. So we see that $M \trianglelefteq P$.
But for this case, I have no idea on how to prove $M$ has index $p$.

Answer 1

Theorem 1: Every finite $p$-group is nilpotent.
Proof: see here.

Theorem 2: A finite group is nilpotent if and only if all its maximal proper subgroup are normal.
Proof: see here.

Theorem 3: A proper normal maximal subgroup of a finite group has prime index.
Proof: see here.

Corollary: Every proper maximal subgroup of a $p$-group is normal and has index $p$.

Answer 2

Define a maximal subgroup of a group to be a proper subgroup not contained in any larger great subgroup. They always exist in nontrivial finite groups, since you can begin with the trivial subgroup; if it is maximal, you win, and if it is not then some other proper subgroup contains it; if that is maximal, you win, and if not ... and so on. A strictly increasing chain of proper subgroups must terminate, in a finite group. More generally if the group is nontrivial and "Noetherian" (I made that up) then maximal subgroups exist.

Fix the prime $p$ and induct on the exponent of the $p$-group $G$. Let $M$ be maximal. If $M\supseteq Z(G)$ then, much as you say, it is easy: the quotient map $q:G\to G/Z(G)$ sets up a correspondence between subgroups of $G$ containing $Z(G)$ and subgroups of $G/Z(G)$, and normal subgroups correspond with normal subgroups via $q$; as $Z(G)$ is famously nontrivial in a $p$-group, the inductive hypothesis applies and $M/Z(G)$ is maximal thus normal, thus $M$ is normal (both by the correspondence) and moreover: $$[G:M]=|G|/|M|=(|G|/|Z(G)|)/(|M|/|Z(G)|)=[G/Z(G):M/Z(G)]=p$$As desired.

If $M$ does not contain the centre, $M\cap Z(G)$ is a proper normal subgroup of the Abelian $p$-group $Z(G)$ and $Z(G)/M\cap Z(G)$ is thus a $p$-group. Famously, ("Cauchy's theorem") it must then have an order $p$ element. Let $\zeta\in Z(G)$ represent this order $p$ element; then $\zeta^p\in M$ and $p$ is the smallest positive integer for which this is true.

Cosets partition, and $gM=M$ iff. $g\in M$, so it is clear $\{M,\zeta M,\zeta^2M,\cdots,\zeta^{p-1}M\}$ is a set of $p$ distinct cosets. As $\zeta^p\in M$ and $\zeta$ commutes with every element of $M$, $\langle\zeta\rangle\cdot M=M\cdot\langle\zeta\rangle=\bigsqcup_{j=0}^{p-1}\zeta^jM$ is a subgroup of $G$ (it is well-known and easy to show that for $H,K\le G$, $HK$ is a subgroup of $G$ if one assumes $HK=KH$).

$M$ is a proper subgroup of $\langle\zeta\rangle\cdot M$ so by maximality of $M$ we must have $\langle\zeta\rangle\cdot M=G$ - then $G$ is the union of those $p$ cosets, so $[G:M]=p$ as desired. $M$ is moreover normal, since we know the generic element $g\in G$ is of the form $\zeta^j\cdot m$ for some integer $j$, some $m\in M$ - where, recall, $\zeta$ commutes with every element of $G$ - so if $x\in M$, $gxg^{-1}=\zeta^jmxm^{-1}\zeta^{-j}=\zeta^j\zeta^{-j}\cdot mxm^{-1}=mxm^{-1}\in M$ as required.

Maximal subgroups of a $p$-group always exist, and are always normal of index $p$.

Answer 3

I think the proof by induction is a good idea. But perhaps you should induct on the exponent of the group. So, if $|P|=p^1,$ then the problem holds. Assume true for $|P|=p^n,$ then for $|P|=p^{n+1},$ we have that the center of $P$ (which we can call $Z$) is non-trivial (and also a $p$ group) so $Z$ contains an element of order $p$, call it $x$. But since $Z$ is characteristic and abelion, then the group $X=<x>$ is normal in $P$. Thus $|P/X|=p^{n-1}.$ So any maximal subgroup of the quotient group is normal by the inductive hypothesis. I think from here the rest of the proof is pretty straight forward.