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Dec 17, 2012 at 0:29 vote accept tj_
Dec 16, 2012 at 1:47 history edited John Palmieri CC BY-SA 3.0
fixed answer to question 2; added 33 characters in body
Dec 16, 2012 at 1:46 comment added John Palmieri @Craig Westerland: Oh, you're right. fixed.
Dec 16, 2012 at 1:26 comment added Craig Westerland Actually, the map $g \mapsto (g, 1)$ induces a product which carries $\alpha \otimes \beta$ to $\alpha$ only when $\beta \in H^0(G)$; otherwise the "product" is $0$.
Dec 16, 2012 at 0:58 history answered John Palmieri CC BY-SA 3.0