# 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 |