Regular finite principal series are irreducible

If $\chi$ is regular then $W_{\chi }=1$ by [F1], so ${\dim }_{\mathbb{C}}{\mathrm{End}}_G(I(\chi ))=1$ by [F1], and therefore ${\mathrm{End}}_G(I(\chi ))=\mathbb{C}\cdot \mathrm{id}$: a one-dimensional complex subspace of the endomorphism algebra containing the nonzero element $\mathrm{id}$.

Conversely assume that $\chi$ is not regular, so $W_{\chi }\mathrm{\ne }1$ by [F1] and ${\dim }_{\mathbb{C}}{\mathrm{End}}_G(I(\chi ))=\mathrm{\mid }{W}_{\chi }\mathrm{\mid }>1$ by [F1]. If $I(\chi )$ were irreducible, then ${\mathrm{End}}_G(I(\chi ))$ would be a division ring by [F3], and being finite-dimensional over $\mathbb{C}$ it would equal $\mathbb{C}\cdot \mathrm{id}$ by the eigenvalue argument of [F3]; its dimension would be $1$, contradicting $\mathrm{\mid }{W}_{\chi }\mathrm{\mid }>1$. Hence $I(\chi )$ is not irreducible, and since it is nonzero it has a nonzero proper submodule, that is, it is reducible.

Assume $\chi$ regular. The module $I(\chi )$ is nonzero of dimension $[G:B]\ge 1$ by [F4] and semisimple by [F2]. If it were not simple, [F2] would produce a decomposition $I(\chi )=A\oplus B$ with $A,B\mathrm{\ne }0$, and the projection onto $A$ along $B$ would be an endomorphism $p$ with $p\mathrm{\ne }0$ and $p\mathrm{\ne }\mathrm{id}$, so that $p\mathrm{\notin }\mathbb{C}\cdot \mathrm{id}$; this contradicts step 1.1. Hence $I(\chi )$ is irreducible, with endomorphism algebra $\mathbb{C}\cdot \mathrm{id}$.

Steps 2.1 and 1.2 prove both directions of the equivalence $I(\chi )\text{ irreducible}⟺W_{\chi }=1$, and $W_{\chi }=1⟺\chi \text{ regular}$ is the definition of regularity in [F1]; the dimension is [F4], and [F1] also gives $I(\chi )\cong I(w\cdot \chi )$, so the isomorphism class of a regular principal series depends only on the $S_n$-orbit of $\chi$. The argument used Maschke, Schur and the finite-dimensional eigenvalue principle only, so no choice principle is used.