Euler Phi Function is Multiplicative
Theorem
The Euler $\phi$ function is a multiplicative function:
- $m \perp n \implies \map \phi {m n} = \map \phi m \map \phi n$
where $m, n \in \Z_{>0}$.
Proof 1
Let $R = \set {r_1, r_2, \ldots, r_{\map \phi m} }$ and $S = \set {s_1, s_2, \ldots, s_{\map \phi n} }$ be the reduced residue systems for the respective moduli $m$ and $n$.
We are to show that the set of $\map \phi m \map \phi n$ integers:
- $T = \set {n r + m s: r \in R, s \in S}$
is a reduced residue system for modulus $m n$.
We need to establish the following:
- Each integer in $T$ is prime to $m n$
- No two integers in $T$ is congruent modulo $m n$
- Each integer prime to $m n$ is congruent modulo $m n$ to one of these integers in $T$.
We prove each in turn:
Aiming for a contradiction, suppose $p$ is a prime divisor of $\gcd \set {n r + m s, m n}$ where $r \in R, s \in S$.
We have that $p$ divides $m n$ but $m \perp n$.
Hence from Divisors of Product of Coprime Integers, $p$ either divides $m$ or $n$ but not both.
Without loss of generality, suppose $p \divides m$.
Then as $p \divides n r + m s$, we have $p \divides n r$ and hence $p \divides r$.
But then $p \divides \gcd \set {m, r} = 1$ which is a contradiction.
Similarly if $p \divides n$.
So there is no such prime and hence $n r + m s \perp m n$.
$\Box$
Let $n r + m s = n r' + m s' \pmod {m n}$, where $r, r' \in R, s, s' \in S$.
Then:
- $n \paren {r - r'} + m \paren {s - s'} = k \paren {m n}$ for some $k \in \Z$.
As $m$ divides two of these terms it must divide the third, so:
- $m \divides n \paren {r - r'}$
Now $m \perp n$ so by Euclid's Lemma:
- $m \divides \paren {r - r'}$
or:
- $r \equiv r' \pmod m$
But $r$ and $r'$ are part of the same reduced residue system modulo $m$, so:
- $r = r'$
Similarly for $n$: we get:
- $s = s'$
Hence distinct elements of $T$ can not be congruent modulo $m n$.
$\Box$
Let $k \in \Z: k \perp m n$.
Since $m \perp n$, from Set of Integer Combinations equals Set of Multiples of GCD we can write $k = n r' + m s'$ for some $r', s' \in \Z$.
Suppose there exists some prime number $p$ such that $p \divides m$ and $p \divides r'$.
Such a prime would be a common divisor of both $k$ and $m n$, contradicting $k \perp m n$.
Hence $r' \perp m$ and so is congruent modulo $m $ to one of these integers in $R$.
By the same argument, $s' \perp n$ and so is congruent modulo $n$ to one of these integers in $S$.
Writing $r' = r + a m, \, s' = s + b n$ we have:
- $k = n r' + m s' = n r + m s + m n \paren {a + b} \equiv n r + m s \pmod {m n}$
Hence the result.
$\blacksquare$
Proof 2
From Euler Phi Function of Product, we have:
- $\map \phi {m n} = \map \phi m \map \phi n \paren {\dfrac {\map \gcd {m, n} } {\map \phi {\map \gcd {m, n} } } }$
From hypothesis, we have:
- $\map \gcd {m, n} = 1$
From Euler Phi Function of 1, we have $\map \phi 1 = 1$, giving the result.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $30$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): phi function (totient function)
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): arithmetic function
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Euler's phi function (phi function, totient function)
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- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 25$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 2.3$: Congruences