An invariant irreducible normal representation yields projective inertia operators

For $t\in T$ the conjugate ${}^t\rho$ is a representation of $tN{t}^{-1}=N$ whose character is ${}^t\theta =\theta ={\chi }_{\rho }$ by [F1] and [F2]; hence ${}^t\rho \cong \rho$ by [F3]. Choose $T_t\in \mathrm{GL}(S)$ with $T_t\rho (n)=\rho (tn{t}^{-1})T_t$ for every $n\in N$, and set $T_1:={\mathrm{id}}_S$, which satisfies this relation.

By [A1] write each $i\in I$ as $i=n_i{t}_i$ with $n_i\in N$, $t_i\in T$, and set $P(i):=\rho (n_i)T_{t_i}$. This is well defined because the pair $(n_i,t_i)$ is unique, and $P(1)=\rho (1)T_1={\mathrm{id}}_S$, while $P(n)=\rho (n)T_1=\rho (n)$ for $n\in N$; each $P(i)$ is invertible, with $P(i{)}^{-1}=T_{t_i}^{-1}\rho (n_i{)}^{-1}$.

For $x\in N$ and $i\in I$ one has $P(xi)=\rho (x)P(i)$ and $P(ix)=P(i)\rho (x)$. Indeed $xi=(x{n}_i)t_i$ and $ix=n_i(t_{i}x{t}_i^{-1})t_i$ are the decompositions of [A1], so $P(xi)=\rho (x{n}_i)T_{t_i}=\rho (x)P(i)$, and $P(ix)=\rho (n_i)\rho (t_{i}x{t}_i^{-1})T_{t_i}=\rho (n_i)T_{t_i}\rho (x)=P(i)\rho (x)$ using the intertwining relation of step 1.1.

For every $i\in I$ and $m\in N$ one has $P(i)\rho (m)P(i{)}^{-1}=\rho (im{i}^{-1})$. Indeed $P(i)\rho (m)P(i{)}^{-1}=\rho (n_i)T_{t_i}\rho (m)T_{t_i}^{-1}\rho (n_i{)}^{-1}=\rho (n_i)\rho (t_{i}m{t}_i^{-1})\rho (n_i{)}^{-1}=\rho (im{i}^{-1})$, using $i=n_i{t}_i$ and step 1.1.

For $i,j\in I$ the operator $A(i,j):=P(i)P(j)P(ij{)}^{-1}$ is invertible and satisfies $A(i,j)\rho (m)=P(i)P(j)\rho ((ij{)}^{-1}m(ij))P(ij{)}^{-1}=\rho (ij(ij{)}^{-1}m(ij)j^{-1}{i}^{-1})A(i,j)=\rho (m)A(i,j)$ for every $m\in N$, by step 3.2 applied to $i$, $j$ and $ij$; since $\rho$ is irreducible over the algebraically closed field $\mathbb{C}$, [F4] gives a unique $\alpha (i,j)\in {\mathbb{C}}^{\times }$ with $A(i,j)=\alpha (i,j){\mathrm{id}}_S$, the scalar being nonzero because $A(i,j)$ is invertible, so $P(i)P(j)=\alpha (i,j)P(ij)$ for all $i,j\in I$.

The scalar $\alpha (i,j)$ of step 4.1 depends only on the cosets $iN$ and $jN$: the general element of $iN$ is $ix$ and the general element of $jN$ is $yj$ with $x,y\in N$, so it suffices to compare $A(ix,j)$ and $A(i,yj)$ with $A(i,j)$. For the first, $ixj=(ij)(j^{-1}xj)$ with $j^{-1}xj\in N$, so $P(ixj)=P(ij)\rho (j^{-1}xj)$ by step 3.1 and $\rho (j^{-1}xj)=P(j{)}^{-1}\rho (x)P(j)$ by step 3.2, whence $A(ix,j)=P(i)\rho (x)P(j)\rho (j^{-1}xj{)}^{-1}P(ij{)}^{-1}=P(i)\rho (x)\rho (x{)}^{-1}P(j)P(ij{)}^{-1}=A(i,j)$. For the second, $yj=y\cdot j$ and $iyj=(iy{i}^{-1})(ij)$ with $iy{i}^{-1}\in N$, so $P(yj)=\rho (y)P(j)$ and $P(iyj)=\rho (iy{i}^{-1})P(ij)$ by step 3.1, while $P(i)\rho (y)P(i{)}^{-1}=\rho (iy{i}^{-1})$ by step 3.2, hence $A(i,yj)=P(i)\rho (y)P(j)P(ij{)}^{-1}\rho (iy{i}^{-1}{)}^{-1}=\rho (iy{i}^{-1})A(i,j)\rho (iy{i}^{-1}{)}^{-1}=A(i,j)$, the last equality because $A(i,j)$ is a scalar.

Write $\alpha (iN,jN):=\alpha (i,j)$, which is well defined by step 5.1, so that $P(i)P(j)=\alpha (iN,jN)P(ij)$ for all $i,j\in I$; and $\alpha (N,qN)=\alpha (qN,N)=1$ for every $q\in I$, since $P(1)P(j)=P(j)$ and $P(i)P(1)=P(i)$ with $P(1)={\mathrm{id}}_S$ by step 2.1 force $A(1,j)={\mathrm{id}}_S=A(i,1)$.

The function $\alpha :Q\times Q\to {\mathbb{C}}^{\times }$ is a normalized two-cocycle on $Q=I\mathrm{/}N$ in the sense of [F6]: the normalization is step 6.1, and associativity of composition in $\mathrm{GL}(S)$ gives $\alpha (i,j)\alpha (ij,k)P(ijk)=(P(i)P(j))P(k)=P(i)(P(j)P(k))=\alpha (j,k)\alpha (i,jk)P(ijk)$, while $P(ijk)$ is invertible by step 2.1, so $\alpha (iN,jN)\alpha (ijN,kN)=\alpha (jN,kN)\alpha (iN,jkN)$; passing to cosets, this is the cocycle identity of [F6] on $Q$.

Now let $P^{\mathrm{\prime }}$ be a second family with the same normalization and the same three identities. For each $i\in I$, the operator $C(i):=P(i{)}^{-1}{P}^{\mathrm{\prime }}(i)$ commutes with $\rho (N)$: by step 3.2, which applies to $P^{\mathrm{\prime }}$ by the same computation, $P(i)\rho (m)P(i{)}^{-1}=\rho (im{i}^{-1})=P^{\mathrm{\prime }}(i)\rho (m)P^{\mathrm{\prime }}(i{)}^{-1}$, hence $C(i)\rho (m)=\rho (m)C(i)$ for all $m\in N$. By [F4] there is a unique $c(i)\in {\mathbb{C}}^{\times }$ with $C(i)=c(i){\mathrm{id}}_S$, so $P^{\mathrm{\prime }}(i)=c(i)P(i)$. Step 3.1 for both families gives $c(xi)=c(i)=c(ix)$ for $x\in N$, so $c$ is constant on both left and right $N$-cosets and defines $c:Q\to {\mathbb{C}}^{\times }$, and $c(N)=1$ because $P^{\mathrm{\prime }}(1)=P(1)={\mathrm{id}}_S$. Finally $P^{\mathrm{\prime }}(i)P^{\mathrm{\prime }}(j)=c(i)c(j)\alpha (iN,jN)P(ij)=c(i)c(j)c(ij{)}^{-1}\alpha (iN,jN)P^{\mathrm{\prime }}(ij)$, so the factor set of $P^{\mathrm{\prime }}$ is ${\alpha }^{\mathrm{\prime }}(q,r)=c(q)c(r)c(qr{)}^{-1}\alpha (q,r)$.

Steps 2.1, 3.1, 6.1 and 7.1 produce operators $P(i)\in \mathrm{GL}(S)$ with $P(1)={\mathrm{id}}_S$, $P(n)=\rho (n)$, $P(ni)=\rho (n)P(i)$, $P(in)=P(i)\rho (n)$ and $P(i)P(j)=\alpha (iN,jN)P(ij)$ for the normalized two-cocycle $\alpha$ on $I\mathrm{/}N$. Composing $\alpha$ with the coset projection $I\times I\to (I\mathrm{/}N)\times (I\mathrm{/}N)$ therefore expresses $P$ as a normalized projective representation of $I$ in the sense of [F5] whose factor set is carried by $I\mathrm{/}N$, and step 7.2 shows that every other normalized family differs from $P$ by a ${\mathbb{C}}^{\times }$-valued cochain $c$ on $I\mathrm{/}N$ with ${\alpha }^{\mathrm{\prime }}=c(q)c(r)c(qr{)}^{-1}\alpha$.