Let $G$ be an abelian group ; let $R=\oplus _{g \in G} R_g$ be a commutative unital $G$-graded ring . For a subgroup $H$ of $G$ we denote by $R_H$ the $H$ graded ring as $R_H = \oplus _{h \in H} R_h$ , where $R_h$ is the $h$-th homogeneous part of the $G$-graded ring $R$ and by $R_{G/H}$ we denote the $G/H$-graded ring $R_{G/H}=\oplus_{g+H \in G/H} (\oplus_{h \in H} R_{g+h})$ . Now a $G$(abelian) -graded ring $R$ is called strongly graded if $R_gR_{g'}=R_{g+g'} , \forall g, g' \in G$ . My question is the following : Let $G$ be an abelian group , and $H$ be a subgroup of $G$ . Let $R$ be a commutative unital $G$-graded ring such that $R_H$ and $R_{G/H}$ are strongly graded . Then is $R$ strongly graded ?
If it is not true in general , then is there any extra condition on $G$ or on the subgroup $H$ or on the ring $R$ which forces the assertion to be true ? Has there been any work done on this ?
Suppose $R_H$ and $R_{G/H}$ are strongly graded. Let $a,b\in G$, $a+b=c$, and $x\in R_c$. Since $R_{G/H}$ is strongly graded, there exist $h_i,h_i'\in H$ $y_i\in R_{a+h_i}$ and $z_i\in R_{b+h_i'}$ such that $x=\sum_i y_iz_i$. Note that then we may discard those $y_i$ and $z_i$ such that $h_i'\neq -h_i$, since then $y_iz_i$ is homogeneous of degree different from $c$ and so all such terms must cancel for the sum to be equal to $x$. So we may assume $y_i\in R_{a+h_i}$ and $z_i\in R_{b-h_i}$ for each $i$.
Now since $R_H$ is strongly graded, for each $i$, there exist $u_{ij}\in R_{h_i}$ and $v_{ij}\in R_{-h_i}$ such that $\sum_j u_{ij}v_{ij}=1$. We then have $$x=\sum_i y_iz_i\sum_ju_{ij}v_{ij}=\sum_{i,j}(y_iv_{ij})(z_iu_{ij}).$$ Since $y_iv_{ij}\in R_a$ and $z_iu_{ij}\in R_b$, this shows $x\in R_aR_b$, so $R$ is strongly graded.