The order of all the elements of a finite additive group of a ring divides characteristic Let $(R,+,\cdot)$ be any ring, $|R|<\infty$.
We know that  $\forall a\in (R,+),\, ord(a)<\infty$ .
1) How can we prove that: $$\forall r\in(R,+,\cdot),\, ord(r)|Char(R)  \ \ \ \ (\dagger)$$ 
2) Can we claim from $ (\dagger)$ that $Char(R) \neq 0$ ?
Note: $Char(R)$ is the Characteric of the ring $R$.
 A: A ring with characteristic $0$ is necessarily infinite, because it contains a subring isomorphic to the integers.
So if a ring is finite, it has nonzero characteristic.
Then it's basic group theory: if $G,+$ is a group and $kg=0$, for $g\in G$ and $k>0$ an integer, then $\operatorname{ord}(g)\mid k$.
Since $kr=0$, for every $r\in R$, where $k$ is the characteristic of $R$, you're done.
A: I'm going to prove $\text{char} \ R \neq 0$ first and then prove your first statement because to me, that makes more sense.


*

*$\text{char} \ R \neq 0$


To prove this, we must simply prove that there is some integer $n$ such that $n \cdot a=0$ for all $a \in R$. It is given that $\text{ord} \ a$ is finite for all $a \in R$. We are also given that $R$ is finite. Therefore, we can look at all of the finite orders and find a least common multiple of the orders called $n$. This means that for any element $a$, $\text{ord} \ a \mid n$, so $n=t \ \text{ord} \ a$ for some $t \in \Bbb{N}$. Thus:
$$n \cdot a=(t \ \text{ord} \ a) \cdot a=t(\text{ord} \ a \cdot a)=t \cdot 0=0$$
The following shows $n \cdot a=0$ for all $a \in R$, concluding the proof.


*$\text{ord} \ a \mid \text{char} \ R$ for all $a \in R$


Let $m=\text{ord} \ a$ and $n=\text{char} \ R$. By the Division Theorem, there exists $q, r \in \Bbb{N}$ such that $0 \leq r < m$ and $n=qm+r$. In order to prove that $m \mid n$, we will prove $r=0$. By our previous equation, $r=n-qm$. We will now examine $r \cdot a$:
$$r \cdot a=(n-qm) \cdot a=n \cdot a-qm \cdot a=0-q \cdot 0=0$$
Thus, $r \cdot a=0$. However, $m$ is the least positive integer such that $m \cdot a=0$, yet $r < m$. Therefore, $r$ can not be positive, so $r=0$. Thus, $n=qm+r=qm$, meaning $m \mid n$, or, in other words, $\text{ord} \ a \mid \text{char} \ R$.
