$g(m)$ is a multiplicative fuction.

Question

$g(m)=\sum_{n=1, (n,m)=1}^{n=m} e^{2\pi i n /m}$

I tried but didnt able to show it. Also how to show that $g(p)=-1$ for $p$ prime. I tried it just by splitting the series and collecting the terms but that doesn’t worked out!!!

Answer 1

Introducing

$$g(n) = \sum_{q=1, (n,q)=1}^n e^{2\pi i q/n}$$

we observe that with an Iverson bracket

$$[[n=1]] = \sum_{q=1}^n e^{2\pi i q/n} = \sum_{d|n} \sum_{q=1, (n,q)=d}^n e^{2\pi i q/n} = \sum_{d|n} \sum_{p=1, (n,pd)=d}^{n/d} e^{2\pi i pd/n} \\ = \sum_{d|n} \sum_{p=1, (n/d,p)=1}^{n/d} e^{2\pi i p/(n/d)} = \sum_{d|n} g(n/d) = \sum_{d|n} g(d).$$

We thus have by Mobius inversion that

$$g(n) = \sum_{d|n} [[d=1]] \mu(n/d) = \mu(n).$$

Now to see that $\mu$ is multiplicative we want to show that $\mu(mn) = \mu(m)\mu(n)$ when $(m,n)=1.$ There are two cases, at least one of $m$ and $n$ is not squarefree or both are squarefree. Let $P$ and $Q$ be the primes that divide $m$ and $n$ respectively, so that $P\cap Q=\emptyset.$ For the first case $\mu(mn)$ is zero because the primes with exponent at least two in the factorizations of $m$ and/or $n$ appear unchanged in $mn,$ which is therefore not squarefree either. We also have zero for $\mu(m)\mu(n)$ because at least one of these is not squarefree. This leaves the case of $m$ and $n$ being squarefree. We get for $\mu(mn)$ the value $(-1)^{|P|+|Q|}$ and for $\mu(m)\mu(n) = (-1)^{|P|} (-1)^{|Q|}$ and we have agreement here as well, proving the claim.