Right now I'm studying out of Audrey Terras' book Fourier Analysis on Finite Groups and Applications and we're on the section where we're talking about $(\mathbb{Z}/n\mathbb{Z})^*$ and when this group is cyclic. I understand that when $n=2,4,p,p^e,2p^e$, then $(\mathbb{Z}/n\mathbb{Z})^*$ is cyclic (where $p$ is an odd prime and $(\mathbb{Z}/n\mathbb{Z})^*$ is the multiplicative group mod n such that it has an inverse).
Really my ultimate goal is to show that the sum of all primitive roots mod p add up to the Mobius function $\mu(p-1)\bmod p$, but I can't wrap my head around the connection between cyclic groups and the primitive roots. Can anyone help me?
For prime $p$ : A primitive root mod $p$ is any $x$ with $1\leq x<p$ such that the least positive integer $n$ for which $x^n\equiv 1\pmod p$ is $n=p-1.$ Let $G$ be the group $G=(Z/Zp)^*=\{1,...,p-1\}$ with multiplication mod $p.$ Then $x$ is a primitive root mod $p$ iff $x$ is a generator for $G.$ That is, in $G$ we have $\{x^n :0\leq n<p\}=G.$ If $x$ is a primitive root mod $p$ then the set of all primitive roots can be represented in $G$ (that is, reduced mod $p$) as $\{x^n :1\leq n<p \land \gcd (n,p-1)=1\}$. A finite group $H$ with $|H|$ members is defined to be cylic iff some $x\in H$ is a generator for $H$, that is $H=\{x^n :1\leq n\leq |H|\}$. The existence of a primitive root mod $p$ is a special case of the following with $F=Z/Zp$.
Lemma: If $F$ is a finite field then the multiplicative group $G=F\backslash \{0\}$ is cyclic.