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@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.
@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
