proof of the Burnside basis theorem

Let $P$ be a $p$-group and $\mathrm{\Phi }(P)$ its Frattini subgroup.

Every maximal subgroup $Q$ of $P$ is of index $p$ in $P$ and is therefore normal in $P$. Thus $P/Q\cong {\mathbb{Z}}_p$. So given $g\in P$, $g^p\in Q$ which proves $P^p\le Q$. Likewise, ${\mathbb{Z}}_p$ is abelian so $[P,P]\le Q$. As $Q$ is any maximal subgroup, it follows $[P,P]$ and $P^p$ lie in $\mathrm{\Phi }(P)$.

Now both $[P,P]$ and $P^p$ are characteristic subgroups of $P$ so in particular $F=[P,P]P^p$ is normal in $P$. If we pass to $V=P/F$ we find that $V$ is abelian and every element has order $p$ – that is, $V$ is a vector space over ${\mathbb{Z}}_p$. So the maximal subgroups of $P$ are in a 1-1 correspondence with the hyperplanes of $V$. As the intersection of all hyperplanes of a vector space is the origin, it follows the intersection of all maximal subgroups of $P$ is $F$. That is, $[P,P]P^p=\mathrm{\Phi }(P)$.

Title	proof of the Burnside basis theorem
Canonical name	ProofOfTheBurnsideBasisTheorem
Date of creation	2013-03-22 15:46:25
Last modified on	2013-03-22 15:46:25
Owner	Algeboy (12884)
Last modified by	Algeboy (12884)
Numerical id	12
Author	Algeboy (12884)
Entry type	Proof
Classification	msc 20D15