$G$-graded $A$-modules.

For a subgroup $H<G$ with finite index, we fix a complete set of representatives $I\subseteq G$ of $G/H$. Let $$\label{eqn:ringchange}
A^{[H]}:=\bigoplus_{h\in H}(A^{[H]})_h\ \ \ \mbox{where}\ \ \ 
(A^{[H]})_h:=\left( A_{i-j+h} \right)_{i,j\in I}.$$ Then $A^{[H]}$ has a structure of an $H$-graded ring whose multiplication $(A^{[H]})_h\times (A^{[H]})_{h'}\to (A^{[H]})_{h+h'}$ for $h,h'\in H$ is given by $$(a_{i-j+h})_{i,j\in I}\cdot (a'_{i-j+h'})_{i,j\in I}
:=\left(\sum_{k\in I}a_{i-k+h}\cdot a'_{k-j+h'}\right)_{i,j\in I}.$$ It is easy to see that the ring structure of $A^{[H]}$ does not depend on the choice of $I$. Moreover the choice of $I$ does not change the graded structure of $A^{[H]}$ up to graded-Morita equivalence in the following sense.

\[graded Morita\] With the notation above, we have an equivalence of categories: $$\operatorname{\mathsf{Mod}}^GA\simeq \operatorname{\mathsf{Mod}}^HA^{[H]}$$ which induces an equivalence ${\operatorname{\mathsf{mod}}}^GA\simeq {\operatorname{\mathsf{mod}}}^HA^{[H]}$.

Although similar results already exist (e.g. [@H; @Mo]), we include a complete proof due to lack of suitable references for our setting.

For $M=\bigoplus_{g\in G}M_g$ in $\operatorname{\mathsf{Mod}}^GA$, define $FM\in\operatorname{\mathsf{Mod}}^HA^{[H]}$ by $$FM:=\bigoplus_{h\in H}(FM)_h\ \ \ \mbox{where}\ \ \ (FM)_h:=(M_{i+h})_{i\in I}.$$ Thus $FM=M$ as abelian groups, and the action $(A^{[H]})_h\times (FM)_{h'}\stackrel{\rm act.}\longrightarrow
(FM)_{h+h'}$ is given by $$(a_{i-j+h})_{i,j\in I}\cdot (m_{i+h'})_{i\in I}
:=\left(\sum_{k\in I}a_{i-k+h}\cdot m_{k+h'}\right)_{i\in I}.$$

Let $f:M\to N$ be a morphism of abelian groups. Then $f$ is a morphism in $\operatorname{\mathsf{Mod}}^GA$ if 