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Mar 21, 2018 at 7:23 comment added YCor The assumption is not only that the kernel is not $C$, but that the new quotient map does not split. Otherwise there's the trivial example $C_p\times C_p\to C_p$ (with nonfaithful action). So I see little point in removing one nontrivial assumption by the OP to claim the question is trivial, and add another assumption of yours to make it slightly less trivial.
Mar 21, 2018 at 6:15 history edited Glasby CC BY-SA 3.0
changed an \ltimes to a \rtimes and added the word "this"
Mar 21, 2018 at 6:09 history answered Glasby CC BY-SA 3.0