TITLE: Is the group of units of a finite ring cyclic?
QUESTION [19 upvotes]: The group of units of a finite field is cyclic. 
Is it true that the group of units of a finite ring is also cyclic?
If not, where does the ring structure prevents us from
obtaining the result that is true for fields?

REPLY [2 votes]: In the case of a field $K$, any finite subgroup of the multiplicative group is indeed cyclic. Ruffini's theorem holds only for fields (and for domains).  Thus, if $A^\times$ is finite, the equation
$$x^n-1=0$$
holds for $n=|A^\times|$ and $x\in A$.  However, since $(x-a)(x-b)=0$ does not imply that $x=a$ or $x=b$ anymore, it becomes complicated in the case of non-integral rings.
For $n$ a natural number, the ring ${\mathbb Z}/n{\mathbb Z}$ has a cyclic multiplicative group for $n=p^k$ or $n=2p^k$, but e.g. in the case where $n$ has at least two different odd prime divisors then it becomes non-cyclic, simply because
$$I1.\text{  }U({\mathbb Z}/p^{k}{\mathbb Z})={\mathbb Z}/p^{k-1}(p-1){\mathbb Z}$$
and by the Chinese remainder theorem you have
$$I2.\text{  }U({\mathbb Z}/n{\mathbb Z})=\oplus U({\mathbb Z}/p_i^{e_i}{\mathbb Z})=\oplus {\mathbb Z}/p_i^{e_i-1}(p-1){\mathbb Z}.$$
Clearly, the l.c.m. of the orders of the subgroup on the right strictly divides the order of the group, just because there are at least two even numbers in the list.