A coset basis makes a monomial representation monomial matrices

For $t\in T$ define $f_t:G\to \mathbb{C}$ by $f_t(t{h}^{\mathrm{\prime }}):=\lambda (h^{\mathrm{\prime }}{)}^{-1}$ for $h^{\mathrm{\prime }}\in H$, and $f_t(g):=0$ for $g\mathrm{\notin }tH$. This is well defined because a decomposition $g=t{h}^{\mathrm{\prime }}$ with $t\in T$, $h^{\mathrm{\prime }}\in H$ is unique ([F3]), and $f_t$ lies in ${\mathrm{Ind}}_H^G{\mathbb{C}}_{\lambda }$: for $g=t{h}^{\mathrm{\prime }}$ and $h\in H$ one has $gh=t(h^{\mathrm{\prime }}h)$ and $\lambda (h^{\mathrm{\prime }}h{)}^{-1}=\lambda (h{)}^{-1}\lambda (h^{\mathrm{\prime }}{)}^{-1}$, while for $g\mathrm{\notin }tH$ also $gh\mathrm{\notin }tH$ and both sides vanish.

For claim 2 put $L:=\mathbb{C}{v}_1$ and $H:=\{x\in G:x\cdot L=L\}$. This $H$ is a subgroup of $G$ containing $1$, and each $x\in H$ acts on the line $L$ by a nonzero scalar; writing $x\cdot z=\lambda (x)z$ for $z\in L$ defines a group homomorphism $\lambda :H\to {\mathbb{C}}^{\times }$, because the action of $G$ on $V$ is a group action. Thus $\lambda$ is a linear character of $H$ and $L$ is a one-dimensional $H$-module.

For claim 2 let $W$ be the span of all lines $x\cdot L=\mathbb{C}(x\cdot v_1)$ with $x\in G$. Each spanning line is one of the basis lines $\mathbb{C}{v}_i$, because the action permutes the basis lines, so $W\subseteq V$; and $W$ is $G$-invariant with $W\mathrm{\ne }0$, since $y\cdot (x\cdot L)=(yx)\cdot L$ and $L\subseteq W$. Irreducibility of $V$ forces $W=V$, so every basis line equals some $x\cdot L$ and the distinct lines $x\cdot L$, $x\in G$, are exactly the $r$ basis lines.

The function $f_t$ of step 1.1 takes the value $1$ at $t$ and the value $0$ at every other element of $T$; hence ${\mathrm{ev}}_T(f_{t_i})$ is the $i$-th standard basis vector of ${⨁}_{i=1}^n{\mathbb{C}}_{\lambda }$. By [F2] the map ${\mathrm{ev}}_T$ is an isomorphism, so $f_1,\dots ,f_n$ is a basis of ${\mathrm{Ind}}_H^G\lambda$ and $n=[G:H]$.

Fix $x\in G$ and $t\in T$. By [A1] there are unique $t^{\mathrm{\prime }}\in T$ and $h\in H$ with $xt=t^{\mathrm{\prime }}h$. For $s\in T$ the value $(x\cdot f_t)(s)=f_t(x^{-1}s)$ is nonzero exactly when $x^{-1}s\in tH$, that is $s\in xtH=t^{\mathrm{\prime }}H$, hence exactly when $s=t^{\mathrm{\prime }}$. At that point $x^{-1}{t}^{\mathrm{\prime }}=t{h}^{-1}$ by the relation $xt=t^{\mathrm{\prime }}h$, so $(x\cdot f_t)(t^{\mathrm{\prime }})=f_t(t{h}^{-1})=\lambda (h^{-1}{)}^{-1}=\lambda (h)\mathrm{\ne }0$.

For claim 2 choose a left transversal $S$ of the stabilizer $H$ from step 1.2. Such an $S$ exists by choosing one representative from each of the finitely many nonempty cosets in the finite group $G$; no axiom of choice is needed. The coset argument in [F3] applies to this $H$ and $S$. The map $S\to \{x\cdot L:x\in G\}$, $t\mapsto t\cdot L$, is a well-defined bijection: it is well defined because $th\cdot L=t\cdot L$ for $h\in H$; it is injective because $t\cdot L=t^{\mathrm{\prime }}\cdot L$ gives $t^{\mathrm{\prime }-1}t\cdot L=L$, that is $t^{\mathrm{\prime }-1}t\in H$, hence $t^{\mathrm{\prime }}=t$ by [F3]; and it is surjective because every $x\in G$ can be written $x=th$ with $t\in S$, $h\in H$, giving $x\cdot L=t\cdot L$. By step 1.3 the translates $L_t:=t\cdot L$, $t\in S$, are exactly the basis lines, and $V={⨁}_{t\in S}{L}_t$, each $u\in V$ having a unique expression $u={\sum }_{t\in S}t\cdot u_t$ with $u_t\in L$.

Thus the matrix of $x$ in the basis $f_1,\dots ,f_n$ has its only possibly nonzero entry in the column indexed by $t$ at the row indexed by $t^{\mathrm{\prime }}$, where $xt\in t^{\mathrm{\prime }}H$, and that entry equals $\lambda (h)\mathrm{\ne }0$. The map $t\mapsto t^{\mathrm{\prime }}$ is a permutation of $T$ by [A1], so every row also receives exactly one nonzero entry, namely from the unique $t$ with $xt\in t^{\mathrm{\prime }}H$. Hence the matrix is monomial with nonzero entries among the values of $\lambda$, which proves claim 1.

Define $\mathrm{\Psi }:V\to {\mathrm{Ind}}_H^{G}L$ by $\mathrm{\Psi }(v)(th):=h^{-1}\cdot u_t$ for $t\in S$, $h\in H$, where $v={\sum }_{t\in S}t\cdot u_t$ is the unique expression of step 2.3. This is well defined by [F3], it satisfies the covariance law $\mathrm{\Psi }(v)(g{h}^{\mathrm{\prime }})=h^{\mathrm{\prime }-1}\cdot \mathrm{\Psi }(v)(g)$ for $g=th$, $h^{\mathrm{\prime }}\in H$, and so lies in ${\mathrm{Ind}}_H^{G}L$ as in [F1]; the assignment $\mathrm{\Psi }$ is $\mathbb{C}$-linear because the coordinates $u_t$ depend linearly on $v$.

For $f\in {\mathrm{Ind}}_H^{G}L$ put $\mathrm{\Phi }(f):={\sum }_{t\in S}t\cdot f(t)$, an element of $V$ by step 2.3, and $\mathrm{\Phi }$ is $\mathbb{C}$-linear. For $v={\sum }_{t\in S}t\cdot u_t$ step 3.2 gives $\mathrm{\Psi }(v)(t)=u_t$, so $\mathrm{\Phi }(\mathrm{\Psi }(v))=v$. Conversely, for $f\in {\mathrm{Ind}}_H^{G}L$ and $t\in S$ the definition of $\mathrm{\Phi }$ gives $\mathrm{\Psi }(\mathrm{\Phi }(f))(t)=f(t)$, and both $\mathrm{\Psi }(\mathrm{\Phi }(f))$ and $f$ satisfy the covariance law [F1], so they agree on $G$; hence $\mathrm{\Psi }$ is surjective and $\mathrm{\Phi }={\mathrm{\Psi }}^{-1}$.

$\mathrm{\Psi }$ is $G$-equivariant. Let $v={\sum }_{t\in S}t\cdot u_t$ and $x\in G$; for each $t\in S$ write uniquely $xt=t^{\mathrm{\prime }}h$ with $t^{\mathrm{\prime }}\in S$, $h\in H$, so that $x\cdot v={\sum }_t(xt)\cdot u_t={\sum }_t{t}^{\mathrm{\prime }}\cdot (\lambda (h)u_t)$ has $L_{t^{\mathrm{\prime }}}$-component $\lambda (h)u_t$. By step 3.2, $\mathrm{\Psi }(v)(x^{-1}{t}^{\mathrm{\prime }})=\mathrm{\Psi }(v)(t{h}^{-1})=\lambda (h)u_t$, while the function $x\cdot \mathrm{\Psi }(v)$ takes the value $\mathrm{\Psi }(v)(x^{-1}{t}^{\mathrm{\prime }})$ at $t^{\mathrm{\prime }}$ by [F1]; as $t$ runs over $S$ so does $t^{\mathrm{\prime }}$, so $\mathrm{\Psi }(x\cdot v)$ and $x\cdot \mathrm{\Psi }(v)$ agree on $S$ and both satisfy the covariance law, hence they agree on $G$.

Steps 4.1 and 4.2 exhibit $\mathrm{\Psi }:V\to {\mathrm{Ind}}_H^{G}L={\mathrm{Ind}}_H^G\lambda$ as a $G$-equivariant $\mathbb{C}$-linear bijection, where $\lambda$ is the linear character of step 1.2, so $V$ is monomial in the sense of [F5]. Together with step 3.1 this proves both assertions.