TITLE: Crossed product of a C*-algebra by a subgroup
QUESTION [6 upvotes]: Let $A$ be a unital $C^*$-algebra, let $G$ be a compact group, let $\alpha:G\to\mbox{Aut}(A)$ be a continuous action, and let $H$ be a closed subgroup of $G$. Is there any relationship between the crossed products $A\rtimes_\alpha G$ and $A\rtimes_{\alpha|_H}H$?
I really only need this for $G=\mathbb{T}$ the unit circle, and $H=\mathbb{Z}_n$ (identified with the $n$-th roots of unity in $\mathbb{T}$). For these groups, and in the case of the trivial action of the circle on $A$, $A\rtimes \mathbb{Z}_n \cong A\otimes \mathbb{C}^n$ is a corner of $A\rtimes \mathbb{T} \cong A\otimes c_0(\mathbb{Z})$, but I don't know if this is true in general.

REPLY [5 votes]: You always have an injective $*$-homomorphism from $A\rtimes H$ into the multiplier algebra of $A\rtimes G$ (the reason is that you can view functions on $H$ as measures on $G$ which are supported on $H$). If $H$ is open in $G$ (a rather unfrequent situation, as you know), then $A\rtimes H$ sits as a $C^*$-subalgebra in $A\rtimes G$.