The Möbius Function

Definition: The Möbius Function is the arithmetic function $\mu$ defined for all $n \in \mathbb{N}$ by $\mu(n) = \left\{\begin{matrix}0 & \mathrm{if} \: n \: \mathrm{is \: NOT \: square free.}\\ (-1)^{\omega (n)} & \mathrm{if} \: n \: \mathrm{is \: square \: free.} \end{matrix}\right.$.

Here, $\omega (n)$ is the number of distinct prime divisors of $n$. Also, a number is said to be square free if it is not divisible by a square greater than $1$, that is, all of its prime factors have multiplicity $1$.

Below is a table of the first few values of the Möbius function.

$n$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$\mu(n)$	$1$	$-1$	$-1$	$0$	$-1$	$1$	$-1$	$-1$	$0$	$1$

We now establish that the Möbius function is multiplicative.

Lemma 1: If $(m, n) = 1$ then $mn$ is square free if and only if $m$ and $n$ are both square free.

The Lemma is not true in general if $(m, n) \neq 1$. For example, $(2, 6) = 2$, $2$ and $6$ are square free, but $12$ is not square free.

• Proof: Let $m = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ and let $n = q_1^{f_1}q_2^{f_2}...q_t^{f_t}$ be the prime power decompositions of $m$ and $n$ respectively.

• $\Rightarrow$ Suppose that $mn$ is square free. Then $e_i = 1$ and $f_s = 1$ for all $1 \leq i \leq k$ and $1 \leq s \leq t$. Furthermore $p_i \neq q_s$ for all $1 \leq i \leq k$ and for all $1 \leq s \leq t$. So $m$ and $n$ are square free.

• $\Leftarrow$ Suppose that $m$ and $n$ are square free. Since $(m, n) = 1$, $p_i \neq q_s$ for all $1 \leq i \leq k$ and for all $1 \leq s \leq t$. So $mn$ is squarefree. $\blacksquare$

Theorem 2: The Möbius function $\mu$ is multiplicative.

• Proof: Let $(m, n) = 1$. Observe that since $(m, n) = 1$ we have that $mn$ is square free if and only if $m$ and $n$ are square free from Lemma 1. Let $m = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ and $n = q_1^{f_1}q_2^{f_2}...q_t^{f_t}$ be the prime power decompositions of $m$ and $n$ respectively.

• Case 1: Suppose that $mn$ is square free. If $mn = 1$ then $m = 1$ and $n = 1$, and $\mu(mn) = \mu(m)\mu(n)$. So suppose that $mn > 1$.

• Since $mn$ is square free we have that $m$ and $n$ are square free. So $\mu(m) = (-1)^{\omega(m)}$ and $\mu(n) = (-1)^{\omega(n)}$. Since $\omega$ is additive we have that:

(1)

\begin{align} \quad \mu(m) \mu(n) = (-1)^{\omega(m)} (-1)^{\omega(n)} = (-1)^{\omega(m) + \omega(n)} = (-1)^{\omega(m+n)} = \mu(mn) \end{align}

• Case 2: Suppose that $mn$ is not square free. Since $(m, n) = 1$ we must have that at least one of $m$ or $n$ is not square free. Without loss of generality, assume that $m$ is not square free. Then $\mu(mn) = 0$ and $\mu(m) = 0$, hence:

(2)

\begin{align} \quad \mu(mn) = 0 = \mu(m) = \mu(m)\mu(n) \end{align}

• Thus $\mu$ is multiplicative. $\blacksquare$