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Let $m_i > 1$, where $1 ≤ i ≤ n$, be integers, pairwise relatively prime. Let $m = m_1 \cdots m_n$. Let $\phi(m)$ denote the order of the group $(Z/mZ)^×$. The function $\phi : Z_+ → Z_+$ is called the Euler phi function. Show that there exists an isomorphism $(Z/mZ)^× → (Z/m_1Z)^× \times \cdots \times (Z/m_nZ)^×$. In particular, $\phi(m) = \phi(m_1) \cdots \phi(m_n)$.

Attempt: the direction is to use Chinese Remainder Theorem, but I am kinda stuck there at the origin. If mi are all primes, then the result would be immediate, simply by definition. Any hints please?

Leppala
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gter
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  • One way of approaching this is to note that the multiplicative group of the ring of integers $\pmod{n}$ is abelian. Therefore, the structure theorem tells us it can be written as a direct product of cyclic groups exactly as your problem statement describes. – Kaj Hansen Nov 25 '14 at 06:51
  • the following link might be helpful for you: http://math.stackexchange.com/questions/1035701/if-gcda-b-1-then-why-is-the-set-of-invertible-elements-of-mathbb-z-ab – Krish Nov 25 '14 at 07:37

1 Answers1

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You must use the Chinese Remainder Theorem. Note for the Chinese Remainder Theorem, exist an surjective homomorphism from $Z→(Z/m_1Z) × · · · × (Z/m_nZ)$, since $m_iZ$ are coprime ideals. The kernel of this homomorphism will be $\bigcap m_iZ=mZ$, then by the first isomorphism theorem you will have a isomorphism beetwen $Z/Z_m$ and $(Z/m_1Z)\times · · ·\times(Z/m_nZ)$, now taking the units of each side of this isomorphism you conclude the result.

Hugo Rodrigo Mas Ku
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