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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.

user26857
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  • how do you prove it for fields? – Asinomás Jul 25 '16 at 15:59
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    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 $\Bbb R\times\Bbb Z/2\Bbb Z$. – Gregory Grant Jul 25 '16 at 15:59
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    @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 ! –  Jul 25 '16 at 16:02
  • If the characteristic is $p\neq 2$ then then $-1$ is an element of multiplicative order $2$, and no element has additive order $2$. – Asinomás Jul 25 '16 at 16:04
  • @CarryonSmiling : And for ring of characteristic not $2$ , it goes like this : say $f(x)=1 ,f(y)=-1$ , then $f(x)^2=f(y)^2$ implying $f(2x)=f(2y)$ implying $2x=2y$ and char is not $2$ so $x=y$ implying $1=f(x)=f(y)=-1$ contradicting char is not $2$ –  Jul 25 '16 at 16:05
  • @CameronWilliams : Oh well ... but what about the remaining part ? :p –  Jul 25 '16 at 16:06
  • Why do we have $(f(1)-1)^2=f(1)^2+1$ ? – Asinomás Jul 25 '16 at 16:08
  • @CarryonSmiling: A ring of characteristic not $2$ can have an element of additive order 2. $\mathbf{Z} \times \mathbf{F}_2$, for example, has characteristic zero. –  Jul 25 '16 at 16:09
  • O right, that only works for $p>0$ :( – Asinomás Jul 25 '16 at 16:10
  • Also, $\mathbf{F}_3 \times \mathbf{F}_2$ is a ring of characteristic 6, and $(\mathbf{Z} / 4 \mathbf{Z}) \times \mathbf{F}_2$ is a ring of characteristic 4. –  Jul 25 '16 at 16:10
  • Oh, then it doesn't work at all :( (well, for prime characteristic). – Asinomás Jul 25 '16 at 16:11
  • @CarryonSmiling : If a ring has char $p$ a prime then $(a+b)^p=a^p+b^p$ –  Jul 25 '16 at 16:31
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    Related – Daniel Fischer Jul 25 '16 at 17:45
  • @Hurkyl : It might be noted just for interest that If in a commutative unital ring , $2$ is neither $0$ , nor a zero-divisor , then the additive group of the ring is not isomorphic with the group of units . –  Jul 28 '16 at 12:52
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    It's not true for every finite ring: The trivial ring as a counterexample. – principal-ideal-domain Jul 18 '18 at 09:59

2 Answers2

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

user26857
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anon
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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.

  • 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 $U[X]$? –  Jul 25 '16 at 16:21