If $\Lambda \subset \Gamma$ is a subgroup, then $C_{\lambda}^*(\Lambda) \subset C_{\lambda}^*(\Gamma)$ canonically.
I am reading Brown and Ozawa's book . In page-46, it says that right cosets give a direct sum decomposition $$l^2(\Gamma) \cong \oplus l^2(\Lambda g)$$
and hence the left regular representation of $\Gamma$, when restricted to $\Lambda$,is a multiple of the left regular representation of $\Lambda$(multiplicity equals the number of cosets). This implies the claim.
Here $\Gamma$ is a discrete group and $\lambda:\Gamma \to B(l^2(\Gamma))$ denote the left regular representation defined by $\lambda(s)(\delta_t)=\delta_{st}$ for all $s,t \in \Gamma$, where $\{\delta_t: t\in \Gamma\} \subset l^2(\Gamma)$ is the canonical orthonormal basis. I understand that every $g \in \Gamma$ is in $g\Lambda$. I don't understand how the above statements follow and how they imply the claim.
If I assume that $l^2(\Gamma) \cong \oplus l^2(\Lambda g)$, then I was wondering if $B(l^2(\Gamma)) \cong B(\oplus l^2(\Lambda g))$ via the map $\phi$ where $\phi(T)(\oplus\Lambda g)=\oplus \Lambda (Tg)$? Even if this were true, I don't know how this helps my case as well.
Thanks for the help!!