Suppose that there exist a group $K$ and a homomorphism $\mu:G\to K$. Let $$\delta_\mu:C^*(G,P)\to C^*(G,P)\otimes_{\min} C^*(K)$$ be the coaction of Lemma \[faithful coaction exists\]. Then $$\Psi^\mu:=({{\operatorname{id}}}\otimes \tau)\circ\delta_\mu$$ is a conditional expectation of $C^*(G,P)$ with range ${\overline{{\operatorname{span}}}}\{w_pw_q^*: \mu(p)=\mu(q)\}$. If $K$ is an amenable group, then $\Psi^\mu$ is faithful.

Since ${{\operatorname{id}}}\otimes \tau$ and $\delta_\mu$ are linear and norm decreasing, so is $\Psi^\mu$. Since $\Psi^\mu(w_e)=1$ the norm of $\Psi^\mu$ is $1$. We have $$\label{expectation construction eqn}
 \Psi^\mu(w_pw^*_q)=\begin{cases}w_pw^*_q&\text{if $\mu(p)=\mu(q)$}\\0&\text{otherwise,}\end{cases}$$ and hence $\Psi^\mu\circ\Psi^\mu=\Psi^\mu$. Thus $\Psi^\mu$ is a conditional expectation by [@Tomiyama1957].

From we see that $\overline{{\operatorname{span}}}\{w_pw^*_q:\mu(q)=\mu(p)\}\subseteq {\operatorname{range}}\Psi^\mu$. To show the reverse inclusion, fix $b\in {\operatorname{range}}\Psi^\mu$, say $b=\Psi^\mu(a)$ for some $a\in C^*(G,P)$. Also fix $\epsilon>0$. There exists a finite subset $F\subseteq P\times P$ such that $\|a-\sum_{(p,q)\in F}\lambda_{p,q}w_{p}w^*_{q}\|<\epsilon$. Since $\Psi^\mu$ is linear and norm-decreasing, $$\begin{aligned}
\epsilon&>\Big\|a-\sum_{(p,q)\in F}\lambda_{p,q}w_{p}w^*_{q}\Big\|
\geq \Big\|\Psi^\mu\Big(a-\sum_{(p,q)\in F}\lambda_{p,q}w_{p}w^*_{q}\Big)\Big\|\\
&
=\Big\|\Psi^\mu(a)-\Psi^\mu\Big(\sum_{(p,q)\in F}\lambda_{p,q}w_{p}w^*_{q}\Big)\Big\|  =\Big\|b-\sum_{(p,q)\in F\text{, }\mu(p)=\mu(q)}\lambda_{p,q}w_{p}w^*_{q}\Big\|.\end{aligned}$$ Thus $b\in  \overline{{\operatorname{span}}}\{w_pw^*_q:\mu(q)=\mu(p)\}$, and ${\operatorname{range}}\Psi^\mu=\overline{{\operatorname{span}}}\{w_pw^*_q:\mu(q)=\mu(p)\}$.

Now suppose that $K$ is amenable. To see that $\Psi^\mu$ is faithful, we follow the proof of [@LacaRaeburn1996 Lemma 6.5]. Let $a\in  C^*(G,P)$ and suppose that $\Psi^\mu(a^*a)=0$. Let $f$ be an arbitrary state on $C^*(G,P)$. Then $$\begin{aligned}
           0&=f(\Psi^\mu(a^*a))
           =f\circ({{\operatorname{id}}}\otimes \tau)\circ \delta_\mu(a^*a)\\
           &=(f\otimes \tau)\circ \delta_\mu(a^*a)
           =\tau\circ(f\otimes {{\operatorname{id}}})\circ\delta_\mu(a^*a).
          \end{aligned}$$ Since $K$ is amenable, $\tau$ is faithful. Hence $(f\otimes 