# What is the sufficient and necessary conditions that $-1 \in G$, where $G$ is a multiplicative group of a ring.

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 $$\text{Char}(R)=2$$.

• @A.Pongrácz Nice. To obtain the sufficient and necessary condition, can the original conjecture be modified a bit ? – zongxiang yi Dec 31 '18 at 8:43
• Arthur's answer covers that elegantly. – A. Pongrácz Dec 31 '18 at 8:44
• If I were you, I would accept Arthur's answer. That pretty much solves your problem. – A. Pongrácz Dec 31 '18 at 8:58

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 $$2\mid ord(G)$$.
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 $$(-1,-1)$$.