We will proceed in steps. First
Theorem
Given that \(f(n)\) is a multiplicative function, then
$$
\begin{align*}
g(n) = \sum_{d\mid n} f(d)
\end{align*}
$$
is also multiplicative.
Proof
Given integers \(m\) and \(n\). Suppose that \(\gcd(m,n)=1\) and that \(f(n)\) is multiplicative. Define
$$
\begin{align*}
g(n) = \sum_{d \mid n} f(d).
\end{align*}
$$
We will show that (g) is also multiplicative; that is,
$$
\begin{align*}
g(mn) = g(m)g(n)
\end{align*}
$$
Observe that since \(\gcd(n,m) = 1\), then if \(d \mid mn\), \(m\) and \(n\) share no prime factors and so if we have some factor of \(mn\), then it must be a combination of two factors, one from \(m\) and one from \(n\). Thus, we can write \(d = ab\) where \(a \mid m\) and \(b \mid n\). Based on this
$$
\begin{align*}
g(mn) &= \sum_{d|mn} f(d) \\
&= \sum_{a \mid m} \sum_{b \mid n} f(ab) \\
&= \sum_{a \mid m} \sum_{b \mid n} f(a)f(b) \quad \text{(since $f$ is multiplicative)}\\
&= \sum_{a \mid m} f(a) \sum_{b \mid n} f(b) \\
&= g(m)g(n). \ \blacksquare \\
\end{align*}
$$
Next we need the following:
Lemma
$$
\begin{align*}
\sum_{d\mid n} \mu(d) =
\begin{cases}
1 & \text{if } n = 1, \\
0 & \text{if } n > 1
\end{cases}
\end{align*}
$$
Proof
If \(n = 1\), then the only divisor of \(n\) is \(1\). Thus, \(\sum_{d \mid 1} \mu(d) = \mu(1) = 1\) by definition. Otherwise, \(n > 1\). We can show this by induction of the number of different prime factors of \(n\). So suppose that \(n = p^{\alpha}\). Since \(p\) is prime, then the only divisors of \(p^{\alpha}\) are
$$
\begin{align*}
1,p,p^2,\ldots,p^{\alpha}
\end{align*}
$$
Thus,
$$
\begin{align*}
\sum_{d\mid n} \mu(d) &= \mu(1) + \mu(p) + \ldots + \mu(p^{\alpha}) \\
&= 1 - 1 + 0 + \ldots + 0 = 0.
\end{align*}
$$
Now, suppose this is true for at most \(k\) distinct factors. Then we will show it’s true for \(k+1\) distinct factors. Write \(n = sq^{\beta}\) where \(s\) has \(k\) distinct prime factors and \(q\) is prime and \(q \not\mid s\). Then, observe that every divisor \(d\) of \(n\) can be written uniquely as
$$
\begin{align*}
d = aq^{j} \quad \text{ where } \quad a \mid s \quad \text{ and } \quad 0 \leq j \leq \beta
\end{align*}
$$
Furthermore, note that \(\gcd(a,q^j) = 1\) for any \(j\). Therefore,
$$
\begin{align*}
\mu(d) = \mu(aq^{j}) = \mu(a)\mu(q^j).
\end{align*}
$$
Then
$$
\begin{align*}
\sum_{d\mid n} \mu(d) &= \sum_{a \mid s}\sum_{j=0}^{\beta} \mu(aq^j) \\
&= \sum_{a \mid s}\sum_{j=0}^{\beta} \mu(a)\mu(q^j) \quad \text{(Since $\mu$ is multiplicative)} \\
&= \sum_{a \mid s} \mu(a) (\mu(1) + \mu(q) + \ldots + \mu(q^{\beta})) \\
&= \sum_{a \mid s} \mu(a) (1 - 1 + 0 + \ldots + 0) = 0. \\
\end{align*}
$$
Thus, by induction, we see that when \(n > 1\), then \(\sum_{d \mid n} \mu(d) = 0\). \(\ \blacksquare\)
References
- Math310 Class Notes by Matthias Beck