# proof of the Burnside basis theorem

Let $P$ be a $p$-group and $\Phi(P)$ 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 $P^{p}\leq Q$. Likewise, $\mathbb{Z}_{p}$ is abelian so $[P,P]\leq Q$. As $Q$ is any maximal subgroup, it follows $[P,P]$ and $P^{p}$ lie in $\Phi(P)$.

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, $[P,P]P^{p}=\Phi(P)$.

Title proof of the Burnside basis theorem ProofOfTheBurnsideBasisTheorem 2013-03-22 15:46:25 2013-03-22 15:46:25 Algeboy (12884) Algeboy (12884) 12 Algeboy (12884) Proof msc 20D15