It is obvious that every nonzero element in finite field $\mathbb{F}_{2^n}$ has an odd order. So I was wondering if the following conjecture was true or false.
Notations: Let $(R,+,\times)$ be a finite ring with an identity. The characteristic of ring $R$ is denoted by $\text{char}(R)$. The multiplicative order of an invertible element $e$ is denoted by $\text{ord}(e)$. The set of all invertible elements is denoted by $R^\times$.
If $\text{char}(R)=2$, then $\text{ord}(e)\equiv1\pmod{2}$ for any $e\in R^\times$.
For a weaker conjecture, we have
If $\text{char}(R)=2$, then $\text{ord}(e)\not\equiv 0\pmod{4}$ for any $e\in R^\times$.
For $R=\mathbb{F}_{2^n}$, the conjecture is true. However I think characteristic and multiplicative order are two independent matters. Because I know few rings $R$ with $\text{char}(R)=2$. Could anyone prove or disprove any of the two conjectures?
PS: the problem is araised from Here, in which I need to find out all the elements of order $4$.
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Thanks to user26857 and Jyrki Lahtonen. The two conjectures are false.
For any $p$ and $d$, consider the ring $R=\mathbb{F}_p[x]/(x^d-1)$ and the coset $e=x \in R$. Then the multiplicative order of $e$ is $d$.
- Letting $p=2$ and $d=4$, it disproves the first conjectrure.
- Letting $p=2$ and $d=2$, it disproves the second conjectrure.