Let $p$ be a prime number. Let $U(\mathbb{Z}/p\mathbb{Z})$ denote the group of multiplicative units in $\mathbb{Z}/p\mathbb{Z}$. It is well-known that
- $U(\mathbb{Z}/p\mathbb{Z})=\{1,2, ..., p-1\}$ mod $p$
- $U(\mathbb{Z}/p\mathbb{Z})$ is cyclic.
Let $d$ be any divisor of $p-1$ different from $1.$ Then it seems to me that the sum of elements (mod $p$) in $U(\mathbb{Z}/p\mathbb{Z})$ whose multiplicative order divides $d$ is always divisible by $p$. In symbols,
$$\sum_{a \in \{1,2,..., p-1\} \text{ such that } ord(a)|d }a=0 \text{ mod } p$$
However, I have a hard time proving it. In fact, I am not even sure if it is true. I verified this claim for a few primes and it always seems to work. But I have a hard time showing it in general.
Any help with proving (or disproving) the above claim would be greatly appreciated.
Thank you for your time,
Pawel