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?


A quick google search finds the paper Finite Rings Having a Cyclic Multiplicative Group of Units by Robert W. Gilmer, Jr. which settles the question as follows:

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This is complemented by proving that every finite commutative ring is a direct sum of primary rings and that its group of units is the direct product of the group of units of the primary rings, and is cyclic iff each part is cyclic and they have coprime orders.

  • $\begingroup$ Which of these cases covers ${\bf Z}/2p^r{\bf Z}$, $p$ an odd prime? $\endgroup$ – Gerry Myerson Oct 18 '11 at 2:30
  • $\begingroup$ @Gerry, these are not primary rings. Please see the complete first page in the link I gave. $\endgroup$ – lhf Oct 18 '11 at 2:32
  • $\begingroup$ @lhf: Thanks for your answer. I am trying to see, where does the proof of the theorem for finite fields ("the multiplicative subgroup is cyclic") break down, if we raise the condition that we have a field and instead we have a ring. See for example Theorem 1.9, p. 177 in Lang's Algebra. An outline of the proof is as follows: we consider the decomposition of the group of units into the direct sum of its p-subgroups and show that each p-subgroup is cyclic... $\endgroup$ – Manos Oct 18 '11 at 2:50
  • $\begingroup$ ...In particular let $\alpha$ be an element of maximal period in the $p_i$-subgoup, say $p_i^{r_i}$. Then every element of the $p_i$-subgroup is a root of the polynomial $x^{p_i^{r_i}}-1$. Then if the subgroup is not the cyclic group generated by $\alpha$, we have that the polynomial has more than $p_i^{r_i}$ elements, contradiction. $\endgroup$ – Manos Oct 18 '11 at 2:51
  • 3
    $\begingroup$ @Manos: In an arbitrary ring, it is false that a polynomial of degree $n$ has at most $n$ roots. For example, if $n=pq$ with $p$ and $q$ distinct primes, both congruent to $1$ modulo $4$, then $x^2+1$ has $4$ roots: one for each choice of a square root of $-1$ modulo $p$ and choice of a square root of $-1$ modulo $q$. That's where the argument breaks down for arbitrary rings. It also breaks down for division rings ($x^2+1$ has infinitely many solutions in the real quaternions). $\endgroup$ – Arturo Magidin Oct 18 '11 at 3:39

Given any ring $A$ with non-trivial group of units $A^\times$, the ring $A\times A$ will have a non-cyclic group of units.

This is because for rings $A$ and $B$ one has $(A\times B)^\times=A^\times\times B^\times$ and if $G$ is a non-trivial group the group $G\times G$ is never cyclic.

  • $\begingroup$ Nice answer, thanks! $\endgroup$ – Manos Oct 18 '11 at 16:29

No, $\mathbb Z/8$ has unit group $\{1,3,5,7\}$ mod 8 which is a 2,2 group.

In number-theoretic situations there are coherent things that can be said, and/but in general I think nothing decisive can be said.


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


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 $$U({\mathbb Z}/p^{k}{\mathbb Z})={\mathbb Z}/p^{k-1}(p-1){\mathbb Z}$$ and by the Chinese remainder theorem you have

$$U({\mathbb Z}/n{\mathbb Z})=\oplus U({\mathbb Z}/p_i^{e_i}{\mathbb Z})=\oplus {\mathbb Z}/p_i^{e_i-1}(p-1){\mathbb Z}.$$

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.


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