# composition of multiplicative functions

###### Theorem.

If $f$ is a completely multiplicative function and $g$ is a multiplicative function, then $f\circ g$ 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 CompositionOfMultiplicativeFunctions 2013-03-22 16:09:50 2013-03-22 16:09:50 Wkbj79 (1863) Wkbj79 (1863) 6 Wkbj79 (1863) Theorem msc 11A25