- How many $[u]\in(\mathbf{Z}/ab\mathbf{Z})^\ast$ satisfy the equations $u\equiv 1 \bmod \ a$, $u\equiv 1 \bmod \ b$? I somehow believe that the answer might be $(a,b)$. Is this actually true?
- Is the cardinality of the cokernel of the group homomorphism $(\mathbf{Z}/ab\mathbf{Z})^\ast\rightarrow(\mathbf{Z}/a\mathbf{Z})^\ast\times(\mathbf{Z}/b\mathbf{Z})^\ast$ equal to $\phi((a,b))$?
This would solve the related question on the formula for $\phi$. Thanks in advance.
The kernel of $U(ab)\to U(a)\times U(b)$ is comprised of the integer representatives
$$1+[a,b],~~1+2[a,b],~~1+3[a,b],~~\cdots,~~ 1+ab$$
the number of which is $ab/[a,b]=(a,b)$. (This fact can be argued elementarily via the usual "universal properties" argument ($d\mid a,b\Leftrightarrow a,b\mid ab/d\Leftrightarrow [a,b]\mid ab/d\Leftrightarrow d\mid ab/[a,b]$.)
There is a sequence $U(ab)\to U(a)\times U(b)\to U((a,b))$ where the latter homomorphism is given by $(x,y)\mapsto xy^{-1}$, which can be checked is exact by Chinese Remainder Theorem, so the cokernel does indeed have cardinality $\varphi((a,b))$.
But if the kernel of $U(ab)\to U(a)\times U(b)$ has $(a,b)$ elements, so the image has size $\varphi(ab)/(a,b)$, and so this is the size of the kernel of $U(a)\times U(b)\to U((a,b))$, and so the size of $U((a,b))$ is also given by $\varphi(a)\varphi(b)(a,b)/\varphi(ab)$, which rearranged yields the formula in KCd's original MO answer.