Burnside Basis Theorem

For the forward direction, suppose $X$ is minimally generating. By [L1], $\pi (X)$ spans. If $x\in X$ mapped to zero, or if distinct $x,y\in X$ had the same image, deleting one of them would leave the same spanning image and [L1] would make a proper subset generate, contrary to [F1]. Thus $\pi {\mathrm{\mid }}_X$ is injective. Any proper subset of $\pi (X)$ that spanned would likewise pull back to a proper generating subset of $X$, so $\pi (X)$ spans and no proper subset of it does. By [L2] it contains a basis $B$; since $B$ spans, the failure of every proper subset to span forces $B=\pi (X)$, so $\pi (X)$ is itself a basis.

For the reverse direction, suppose $\pi {\mathrm{\mid }}_X$ is a bijection onto a basis. The basis spans, so [L1] gives $⟨X⟩=P$. Removing any $x\in X$ removes its distinct basis vector; the remaining basis vectors do not span, so [L1] says $X\setminus \{x\}$ does not generate. Thus $X$ is minimal by [F1].

Steps 1.1 and 1.2 prove both implications. For $P=1$, the quotient has the empty basis and the empty set is the minimal generating set, so the same statement applies.