composition of multiplicative functions

Theorem.

If $f$ is a completely multiplicative function and $g$ is a multiplicative function, then $f\mathrm{\circ }g$ is a multiplicative function.

Proof.

First note that $(f\circ g)(1)=f(g(1))=f(1)=1$ since both $f$ and $g$ are multiplicative.

Let $a$ and $b$ be relatively prime positive integers. Then

$(f\circ g)(ab)$	$=f(g(ab))$
	$=f(g(a)\cdot g(b))$ since $g$ is multiplicative
	$=f(g(a))f(g(b))$ since $f$ is completely multiplicative
	$=(f\circ g)(a)(f\circ g)(b)$.

∎

Note that the assumption that $f$ is completely multiplicative (as opposed to merely multiplicative) is essential in proving that $f\circ g$ is multiplicative. For instance, $\tau \circ \tau$, where $\tau$ denotes the divisor function, is not multiplicative:

$$(\tau \circ \tau )(2\cdot 3)=(\tau \circ \tau )(6)=\tau (\tau (6))=\tau (4)=3$$

$$(\tau \circ \tau )(2)\cdot (\tau \circ \tau )(3)=\tau (\tau (2))\cdot \tau (\tau (3))=\tau (2)\cdot \tau (2)=2\cdot 2=4$$

Title	composition of multiplicative functions
Canonical name	CompositionOfMultiplicativeFunctions
Date of creation	2013-03-22 16:09:50
Last modified on	2013-03-22 16:09:50
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	6
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 11A25