# Multiplicative groups of fields

Let $G$ be a finite group, and let be a field on which $G$ acts faithfully. Let $C$, be a $\mathbb{Z}G$-lattice, and let $\alpha$ be a one-cocycle from $G$ to $Hom(C,L^{*})$. Let $L[C]$ the group algebra of the multiplicative abelian group $C$, and let $L(C)$ denote its quotient field. Then $G$ acts on $L(C)$ via the actions on $C$ and on $L$, and there is another $G$-action on $L(C)$ twisted by $\alpha$. The field $L(C)$ with the twisted action is denoted by $L_{\alpha}(C)$. For a field $K$, $K^{*}$ denotes its multiplicative group.

Let $L_{\alpha}^{*}C$ denote the subgroup of $L_{\alpha}(C)^{*}$ generated by $L^{*}$ and $C$. $L[C]$ is a unique factorization domain, and the primes in $L[C]$ form a $\mathbb{Z}$-basis for $\dfrac{L_{\alpha}(C)^{*}}{L_{\alpha}^{*}C}$. Since the $G$-action preserves primes up to units, this basis is the disjoint union of $G$-orbits. $\dfrac{L_{\alpha}(C)^{*}}{L_{\alpha}^{*}C}$ is a direct sum, albeit infinite, of transitive permutation lattices, that is lattices of the form $\mathbb{Z}G/H$ for some subgroup $H$ of $G$. Thus $\dfrac{L_{\alpha}(C)^{*}}{L_{\alpha}^{*}C}=\sum_{i\in I}\mathbb{Z}G/H_{i}$ for some infinite set $I$ and some subgroups $H_{i}$ of $G$. Similarly $\dfrac{L(C)^{*}}{L^{*}C}$ is of the same form.

My question: Has anyone studied the conditions on $G$, $C$, and $\alpha$ under which $\dfrac{L_{\alpha}(C)^{*}}{L_{\alpha}^{*}C}\cong\dfrac{L(C)^{*}}{L^{*}C}$ as $G$-modules?

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