Generalization of $K$ being a field, any finite subgroup $G$ of $K^\star$ is cyclic. Statement. Let $K$ be a field, $G$ is a finite subgroup of $K^\star$ where $K^\star$ are group of non-zero elements of $K$. Then $G$ is cyclic. 
This can be proved by the following method. $G$ is abelian group and finite. We can use PID decomposition and this identifies $x\in G$ s.t. $ord(x)=n$ and $\forall y\in G$, $y^n=1$. Then $x^n=1$ has at most $n$ solution in $K$ but all $G$'s elements are solution. Thus we conclude $G=<x>$.
Consider the above method for the following problem.
Show $(Z_{p^n})^{\star}$ is cyclic.
Consider $Z_{p^n}$. The argument is essentially the same as above but replacing $K$ by $(Z_{p^n})^\star$ and using PID decomposition against $(Z_{p^n})^\star$ instead. The rests are the same if I consider the polynomial in $Z_{p^n}[x]$. 
Q1: Is above proof correct? 
Q2: What is the generalization of the statement? It seems that the statement holds for non field as well. 
Q3: What is the necessary algebraic structure to deduce cyclicity of the unit elements? 
 A: A version of the decomposition theorem says that a finite abelian group can be decomposed as a direct product
$$
G=C(m_1)\times C(m_2)\times\dots\times C(m_k)
$$
(where $C(a)$ is the cyclic group of order $a$) with
$$
m_1\mid m_2 \mid \dots \mid m_k
$$
It is clear that, in this case, $x^{m_k}=1$ for all $x\in G$. In the case $G$ is a subgroup of $K^{\star}$, we cannot have $m_k<|G|=n$, because $x^{m_k}=1$ has at most $m_k$ solutions in $K$. Therefore $m_k=n$ and $G=C(n)$.
Note that this cannot be invoked if $G$ is a finite subgroup of the group of units of a ring that is not a domain (or, equivalently, a field).
Indeed, this fails for $R=\mathbb{Z}/8\mathbb{Z}$, where the group of units is the Klein $4$-group that's not cyclic.
A: Actually, one has the following results:


*

*If $p=2$ and $n\ge 2, \enspace \mathbf Z/2^n\mathbf Z^\times\;$ is isomorphic to the direct sum of the subgroups of order $2$ generated by $-1$ and the subgroup of order $2^{n-2}$ generated by $5$.

*If $p>2,\enspace(\mathbf Z/p^n\mathbf Z)^\times\;$ is cyclic.

