Maximal subgroups of a finite $p$-group are the inverse images of Frattini hyperplanes

If $M$ is maximal in $P$, [F1] and [L3] make $M\mathrm{/}\mathrm{\Phi }(P)$ maximal proper in $E$. Choose a basis of this subgroup and extend it by [L2] to a basis of $E$. Maximality permits exactly one added basis vector, since two would create a proper intermediate span. Thus $M\mathrm{/}\mathrm{\Phi }(P)$ has codimension one.

The coordinate of the omitted basis vector defines a nonzero linear homomorphism $E\to \mathbb{Z}\mathrm{/}p$ whose kernel is $M\mathrm{/}\mathrm{\Phi }(P)$. Conversely, let $\lambda :E\to \mathbb{Z}\mathrm{/}p$ be linear and nonzero and choose $v\in E$ with $\lambda (v)=1$. Every $x\in E$ splits as $x=(x-\lambda (x)v)+\lambda (x)v$ with the first summand in $\ker \lambda$, and $v\mathrm{\notin }\ker \lambda$, so a basis of $\ker \lambda$ together with $v$ spans $E$ and is independent; it is therefore a basis of $E$ by [L2], and $\ker \lambda$ has codimension one. Any subgroup strictly between $\ker \lambda$ and $E$ would contain some $y$ with $\lambda (y)\mathrm{\ne }0$ and hence a scalar multiple of $y$ equal to $v$ modulo $\ker \lambda$, so it would be all of $E$; thus $\ker \lambda$ is maximal proper and [L3] makes its inverse image maximal in $P$.

The quotient and inverse-image maps in [L3] are inverse, so steps 1.1 and 2.1 give the stated classification. The trivial group has neither maximal subgroups nor nonzero linear homomorphisms.