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I am trying to prove the following conjecture.

Let $(R, +,\times)$ be a finite ring with an identity. Let $G$ be a subgroup of $(R,\times)$ with order $d$. Then $-1\in G$, if and only if $2\mid d$ or $\text{Char}(R)=2$.

My attempt:

$\Leftarrow$: Case 1:If $\text{Char}(R)=2$, then $-1=1$. Since $G$ is a multiplicative group, the indentity $1\in G$. Hence $-1\in G$.

Case 2: If $2\mid d$, then ...(I do not know how to prove in this case.)

$\Rightarrow$: Suppose that $-1 \in G$. Then $(-1)^2=1\in G$. The multiplicative order of $-1$ is either $2$ or $1$. If $\text{Ord}(-1)=1$, then $-1=1$. Consequently, $\text{Char}(R)=2$. If $\text{Ord}(-1)=2$, then it must have $\text{Ord}(-1) \mid \text{Ord}(G)$, i.e., $2\mid d$.

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So does $2\mid d$ implies that $-1\in G$ ? If the conjecture does not hold, then can we have a sufficient and necessary condition of $-1 \in G$?

Thanks to Arthur, we have the following result using the fact that $-1$ is the only element of order $2$.

Let $(R, +,\times)$ be an integral domain. Let $G$ be a subgroup of $(R,\times)$ with order $d$. Then $-1\in G$, if and only if $2\mid d$ or $\text{Char}(R)=2$.

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    $\begingroup$ @A.Pongrácz Nice. To obtain the sufficient and necessary condition, can the original conjecture be modified a bit ? $\endgroup$ – zongxiang yi Dec 31 '18 at 8:43
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    $\begingroup$ Arthur's answer covers that elegantly. $\endgroup$ – A. Pongrácz Dec 31 '18 at 8:44
  • $\begingroup$ If I were you, I would accept Arthur's answer. That pretty much solves your problem. $\endgroup$ – A. Pongrácz Dec 31 '18 at 8:58
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Counterexample: let $R=\Bbb Z_{12}$ with standard addition and multiplication, and $G=\{1,5\}$.

As for sufficient conditions, if $R$ is an integral domain with characteristic different from $2$, then the equation $x^2=1$ has exactly two solutions (rewrite to $(x-1)(x+1)=0$ and use the definition of integral domains). Thus $-1$ is the only element with multiplicative order $2$, and by Cauchy's theorem must be in $G$ if $2\mid ord(G)$.

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It is clearly not true: let $R:= (\mathbb{Z}_3, +, \cdot)\times (\mathbb{Z}_3, +, \cdot)$. Then $(1,1)$ is the identity element, and the $2$-element subgroup in the multiplicative group $\{(1,-1),(1,1)\}$ does not contain $(-1,-1)$.

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