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I'm trying to prove that Euler's $\phi$ function is multiplicative i.e. $\phi(mn) = \phi(m)\phi(n)$ when m and n are coprime. A key step in the proof is this:

$A = \{a \mid 1 ≤ a < m, \gcd(a, m) = 1\}$

$B = \{b\mid 1 ≤ b < n, \gcd(b, n) = 1\}$

$C = \{c \mid 1 ≤ c < mn, \gcd(c, mn) = 1\}.$

Then we have that |A| = φ(m), |B| = φ(n), and |C| = φ(m × n). I need to show that C has equally many elements as the set A × B = {(a, b) | a ∈ A, b ∈ B}, from which the claim follows. By using the Chinese remainder theorem by which the mapping $\mathbb{Z}_{mn} → \mathbb{Z}_m × \mathbb{Z}_n$, is bijective in case of m and n coprime. Now it should be enough to prove that $A ⊂ \mathbb{Z}_{m}$, $B ⊂ \mathbb{Z}_n$, and $C ⊂ \mathbb{Z}_{mn}$. I'm not able to prove it. Could you clarify what's the relation between the sets A,B, and C and the quotient set $\mathbb{Z}_{x}$?

Bernard
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Michaelangelo Meucci
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2 Answers2

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Notice that in $\mathbb{Z}_m ×\mathbb{Z}_n $ the the elements $(a,b)$ of order $mn$ have to satisfy $gcd(m,a)=gcd(n,b)=1$ ?

How many such elements do we get ? Is that $\phi(m)\phi(n)$ ?

Again in $\mathbb{Z}_{mn}$ , there are $\phi(mn)$ elements are of order $mn$ .

Due to the bijectivity of the map(the one you mentioned), shouldn't these numbers be equal ?

user-492177
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The mapping $\mathbf Z/mn\mathbf Z\longrightarrow \mathbf Z/m\mathbf Z\times \mathbf Z/n\mathbf Z$, is not only bijective, but it is a ring isomorphism, whence an induced group isomorphism of the groups of units: \begin{align} (\mathbf Z/mn\mathbf Z)^\times&\longrightarrow (\mathbf Z/m\mathbf Z)^\times\times(\mathbf Z/n\mathbf Z)^\times, \\ x+mn\mathbf Z&\longmapsto (x+m\mathbf Z,x+n\mathbf Z). \end{align} Now $A, B, C$ are simply the sets of canonical representative of the congruence classes of the multiplicative groups of units in $\mathbf Z/m\mathbf Z, \mathbf Z/n\mathbf Z$ and $\mathbf Z/mn\mathbf Z$ respectively.

Bernard
  • 179,256