Timeline for Questions about proof that all indecomposable module categories over $\operatorname{Rep}(G)$ are equivalent to $\operatorname{Rep}^1(H,\omega)$

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Sep 20, 2023 at 16:05	comment	added	Sean Sanford		This might not be as satisfying, but it is easier to show that indecomposable module categories for $\text{Vec}_G$ are classified by such pairs $(H,\omega)$. From there you can tensor with the invertible $\big(\text{Rep}(G),\text{Vec}_G\big)$-bimodule category $\text{Vec}$ (invertible=Morita equivalence) to find that all indecomposable module categories for $\text{Rep}(G)$ are classified by the same pairs. This is a more modern way of arriving at this classification, but it's certainly less explicit.
Jun 23, 2023 at 20:33	comment	added	shin chan		@LSpice Elements $\omega\in H^2(H,k^)$ are one-to-one with central extensions $1\to k^\to \tilde H\to H\to1$. With $\operatorname{Rep}^1(H,\omega)=\operatorname{Rep}(\tilde H)$ we denote representations of $\tilde H$ such that $k^*$ acts by the trivial character.
Jun 23, 2023 at 19:53	comment	added	LSpice		What is $\tilde H$?
Jun 23, 2023 at 19:46	history	edited	LSpice	CC BY-SA 4.0	Name of article; `\mathrm` -> `\operatorname`
Jun 23, 2023 at 17:57	history	asked	shin chan	CC BY-SA 4.0