Show $\sum_{c\mid d}\psi(c) = d$ where $d \mid p-1$ and $\psi(c)$ is the number of elements in $(\mathbb{Z}/p \mathbb{Z})^{\star}$ that are of order $c$. $p$ is prime.
I am currently reading through Ireland and Rosen's Number Theory text. I have come across this statement, and I haven't been able to wrap my head around it. I would think that because $c \mid d$ and $d \mid p-1$, $\psi(c) = c$, and so the sum $\sum_{c \mid d}\psi(c) > d$.