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Jun 9, 2014 at 14:41 comment added KConrad I meant not just specify the kernel and cokernel as concrete groups, but also specify what the maps to $U({\mathbf Z}/ab{\mathbf Z})$ and from $U({\mathbf Z}/a{\mathbf Z}) \times U({\mathbf Z}/b{\mathbf Z})$ are.
Jun 9, 2014 at 11:10 comment added Karl @KConrad I guess psychologically the kernel has to be $\mathbf{Z}/(a,b)\mathbf{Z}$ and the cokernel $U(\mathbf{Z}/(a,b)\mathbf{Z})$? I agree that I should have written $1$, somehow I am used to modules, hence additive notation.
Jun 9, 2014 at 0:40 comment added KConrad Yes, but you have to make explicit what the kernel and cokernel really are (the map in the middle indeed is the standard map). Also, since these are multiplicative groups throughout, it looks better to use 1 instead of 0 on both ends of the exact sequence.
Jun 8, 2014 at 22:19 comment added Karl @KConrad: as for the "nice 4-term" sequence which you mention, is it $$0\rightarrow \ker\rightarrow U(\mathbf{Z}/ab\mathbf{Z})\rightarrow U(\mathbf{Z}/a\mathbf{Z})\times U(\mathbf{Z}/b\mathbf{Z})\rightarrow \operatorname{Coker}\rightarrow 0$$? +1 for mentioning this very nice proof
Jul 2, 2011 at 1:33 vote accept teil
Apr 18, 2010 at 22:40 comment added KConrad Martin, what argument do you have in mind? A combinatorial argument sounds like something other than a proof by algebra using the formula with prime factorizations.
Apr 18, 2010 at 22:10 comment added Martin Brandenburg "You could prove that by the formula for phi(n) in terms of prime factorizations, but it wouldn't really explain what is going on because it doesn't provide any meaning to the formula." I disagree with this. I think the combinatoric proof of this formula makes it very clear, whereas more abstract formulas of this type should be proven with diagrams.
Apr 18, 2010 at 2:07 history edited KConrad CC BY-SA 2.5
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Apr 18, 2010 at 1:29 history answered KConrad CC BY-SA 2.5