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If $m,n$ are positive integers with g.c.d.$(m,n)=d$ , then we can show by explicitly computing respective totients that $\phi(mn)\phi(d)=\phi(m)\phi(n)d$, I want to know, is there any more elegant way , without explicitly computing using prime factors, to derive this formula, by some bijection or homomorphisms between known algebraic structures, or any other way which doesn't rely on explicit computation? Thanks in advance

EDIT : Some thoughts : Is it easy to show $U_{mn}$ and $U_m \times U_n \times \mathbb (Z_d / U_d)$ have same cardinality ? Or what about $U_{mn} \times U_d$ and $U_m \times U_n \times \mathbb Z_d$ ; or what about $(\mathbb Z_m / U_m ) \times (\mathbb Z_n / U_n)$ and $(\mathbb Z_{mn}/U_{mn}) \times (\mathbb Z_d / U_d)$ ?

Martin Sleziak
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  • This answer might be what you are looking for. – awllower Apr 01 '15 at 12:37
  • @awllower : Apparently , the $P$ in the answer you linked is not exactly g.c.d. but yes the ratio $d/\phi(d)$ is same as $P/ \phi(P)$ , but Ifind the answer too clumsy , could you please reproduce it here , you can freely use Chinese remainder theorem for general rings ... –  Apr 01 '15 at 12:45
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    I will try to find time, to reduce the clumsiness, and then to reproduce here. Thanks for the attention. – awllower Apr 01 '15 at 12:48
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    @awllower : I should thank you . And also if you can find time please take a look at my edit , I have mentioned some pair of groups , if we can show for at least one of these pairs that they are isomorphic ( or at least bijective , having same cardinality ) then we would be done –  Apr 01 '15 at 12:55
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    Possible duplicate of http://math.stackexchange.com/questions/119911/proving-formula-involving-eulers-totient-function. – lhf Apr 01 '15 at 13:46
  • It turns out that almost all the clumsiness and complications come from $P$ defined as the product of all primes dividing both $m$ and $n.$ Once one removes all the appearances of $P$ in the linked answer, after some slight changes my proof is essentially identical to that of the linked answer by Ihf. So I think there is no need of reproducing my clumsy proof here, as your object has actually nothing to do with that mischievous $P,$ and is already explicitly answered, IMO, by Jonas Kiebelbek. – awllower Apr 03 '15 at 08:28

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