The group of units of the cyclic ring of order $p^b$ is cyclic

Question

Let $U(C_{p^b})$ be the group of units of the ring $C_{p^b}$ where $C_{p^b}$ is cyclic of order $p^b$.

Show that $U(C_{p^b})$ is cyclic.

Answer 1

It helps a lot to know that you are reading this paper. (I suggest that you edit that into the question.)

Note that in the paper $p$ is a Fermat prime, hence of the form $2^{2^k}+1$: in particular it is odd, which is relevant to the structure of $U(C_{p^b})$.

For any cyclic group $(C_n,+)$, the automorphisms are given by multiplication by an integer relatively prime to $n$: we get the group $U(n)$ of order Euler's function $\varphi(n)$. Note that if we choose a generator of $C_n$ we get an isomorphism to $(\mathbb{Z}/n\mathbb{Z},+)$ and then the group automorphisms correspond to the elements of the unit group of the ring. So it doesn't really matter whether we think of $(C_n,+)$ as a ring or not; either way we get to the unit group $U(n)$. (But $U(n)$ does not act by ring automorphisms -- these must fix $1$. It's just the action of any unit group of a ring by multiplication on the elements of the ring, which gives automorphisms of the additive group of the ring.)

The structure of $U(n)$ is classical number theory, first determined by Gauss. In particular it is cyclic if and only if $n$ is:
$\bullet$ $1$, $2$ or $4$; or
$\bullet$ An odd prime power $p^b$, or twice an odd prime power $2p^b$.

For a proof see e.g. $\S$ 1.6 of these notes. This answers your question: you have a Fermat -- hence odd -- prime power, so the unit group is cyclic.