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Sep 22, 2014 at 20:08 comment added Puzzled I did not get your example. The action of $\mathbb{Z}/2$ on $B = C$ is not free. What am I missing? Thanks.
Feb 22, 2010 at 0:53 comment added Andrea Ferretti A very nice argument! I have already assigned the right answer mark to Dmitri, who indeed provided a perfectly legitimate example with g=1, but it is very nice to see that the family is never a product for g at least 2.
Feb 20, 2010 at 20:41 history answered Zsolt Patakfalvi CC BY-SA 2.5