Showing that $(\mathbb{Z}/p^{a}\mathbb{Z})^*$ is a cyclic group I'm supposed to show that if $p$ is an odd prime (so its not $2$) that the multiplicative group $(\mathbb{Z}/p^{a}\mathbb{Z})^*$ is cyclic. It seems very hard to do for any $a$, so I'm thinking about using induction on a to make things easy. That being said, I don't really like group theory. How do I show the base case is cyclic and proceed?
Thanks everyone!
 A: For each $d|p-1$ let $X_d$ bet the set of units of $\mathbb Z/p\mathbb Z$ of order $d$, $(\mathbb Z/p\mathbb Z)^\times$ is the disjoint union of each $X_d$.
The elements of $X_d$ are roots of the cyclotomic polynomial $\Phi_d$ of degree $\varphi(d)$, in any field (in this case we use that $(\mathbb Z/p\mathbb Z)^\times$ is a field) a degree $d-1$ polynomial has $\le d-1$ roots.
Therefore $\sum_{d|p-1} \varphi(d) = p-1 = \sum_{d|p-1} |X_d|$ so $|X_d| = \varphi(d)$.
That shows that $(\mathbb Z/p\mathbb Z)^\times$ is cyclic, the next step is to prove $(\mathbb Z/p^2\mathbb Z)^\times$ cyclic by creating a generator for it using the generator of $(\mathbb Z/p\mathbb Z)^\times$. Then you can prove $(\mathbb Z/p^r\mathbb Z)^\times$ cyclic by showing that the generator for $(\mathbb Z/p^2\mathbb Z)^\times$ is also a generator for it.
A: The post marked answer shows the base case and sets up the induction. Show $(Z/p^2Z)^\times$ is cyclic by showing that if $g$ generates $(Z/pZ)^\times$, then $g$ or $g+p$ generates $(Z/p^2Z)^\times)$.
