Let $R$ be a commutative ring with unity, and let $R^{\times}$ be the group of units of $R$. Then is it true that $(R,+)$ and $(R^{\times},\ \cdot)$ are not isomorphic as groups ?

I know that the statement is true in general for fields. And it is trivially true for any finite ring (as $|R^{\times}| \le |R|-1<|R|$, so they are not even bijective).

I can show that the groups are not isomorphic whenever $\operatorname{char} R \ne 2$ , but I am unable to deal with $\operatorname{char} R=2$ case ... Please help. Thanks in advance.

  • $\begingroup$ how do you prove it for fields? $\endgroup$ – Yorch Jul 25 '16 at 15:59
  • 1
    $\begingroup$ How do you prove it for fields? It seems it might happen for some fields, for example it comes close for $\Bbb R$ because $\Bbb R^{\times}$ is isomorphic to $\Bbb R\times\Bbb Z/2\Bbb Z$. $\endgroup$ – Gregory Grant Jul 25 '16 at 15:59
  • 1
    $\begingroup$ @CarryonSmiling : Yeah , that $char R=2$ case is really tricky part and fields do help out a lot , say $f: R \to R^{\times}$ be a group isomorphism if possible , then $f(0)=1$ and also note that $(f(1)-1)^2=f(1)^2+1=f(1+1)-1=f(0)-1=0$ and in a field there is no non-zero zero divisor , so $f(1)=1$ but then injectivity gives $1=0$ , impossible ! $\endgroup$ – user228169 Jul 25 '16 at 16:02
  • 2
    $\begingroup$ Related $\endgroup$ – Daniel Fischer Jul 25 '16 at 17:45
  • 1
    $\begingroup$ It's not true for every finite ring: The trivial ring as a counterexample. $\endgroup$ – principal-ideal-domain Jul 18 '18 at 9:59

Counterexample: $R=\mathbb{R}\times\mathbb{Z}_2$ satisfies $(R,+)\cong(R^\times,\cdot)$.


The conjecture is false. Here is a counterexample.

Suppose $R$ is a ring with the property that every $r \in R^\times$ satisfies $r^2 = 1$.

Then both $(R,+)$ and $(R^\times,\cdot)$ are abelian groups with the property that every element has exponent $2$ — that is, they are vector spaces over $\mathbf{F}_2$.

If $B$ is a basis for a vector space $V$ over $\mathbf{F}_2$, then the elements of $V$ can be identified with finite subsets of $B$. If $B$ is infinite, it has the same cardinality as its set of finite subsets. Consequently, $(R,+)$ and $(R^\times, \cdot)$ are isomorphic if and only $R$ and $R^\times$ have the same cardinality.

Let $X$ be a set of indeterminates, and define the ring

$$ T[X] = \mathbf{F}_2[X] / \langle x^2 - 1 \mid x \in X \rangle $$

$(T[X], +)$ is a vector space whose basis is the set of all finite subsets of $X$. For any $v \in T[X]$, let $\deg(v)$ be the sum of the coefficients of $v$.

For every $v \in T[X]$, $v^2 = \deg(v)$.

Therefore, for every $v \in T[X]$, we either have $v$ is zero divisor ($v^2 = 0$) or $v$ is a unit (with inverse $v$). Thus, $T[X]^\times$ is the set of all elements with $\deg(v) = 1$.

If $X$ is infinite, then $T[X]$ and $T[X]^\times$ have the same cardinality, and therefore $(T[X],+)$ is isomorphic to $(T[X]^\times, \cdot)$ as abelian groups.

  • $\begingroup$ We could also define $U[X] = \mathbf{F}_2[X] / \langle x^2 \mid x \in X \rangle$. Then there is an isomorphism $T[X] \to U[X]$ that sends $x \mapsto x+1$. Maybe it's easier to think about the construction in terms of $U[X]$? $\endgroup$ – user14972 Jul 25 '16 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy