If $R$ is a finite ring, then $\exists_{n>m>0}: x^n=x^m$ for all $x\in R$

Question

I need some help for the following proof:

If $R$ is a finite ring, then $\exists_{n>m>0}: x^n=x^m$ for all $x\in R$.

I feel there's one or more little tricks to use to see how you get to this equation. So far I've noticed that $R$ is finite, so $R^*$, the group of units, is also finite and therefore cyclic and Abelian. We can write $R^*=\{1,y,y^2\dots y^{r-1}\}$ for a generator $y$, if $|R|=r$. We see that for all $x\in R^*$ we have $x^r=1$. I thought maybe you could use this equation.

Additionally, we have that any $t\in R^*$ has $ord(t)|r$

However, I'm still getting stuck, to actually put together an equation that holds for all $x\in R$.

Answer 1

For each $k\in \mathbb N$, define the function $$\begin{align}\text{pow}_k:R&\to R,\\x&\mapsto x^k\end{align}$$ There are finitely many possible functions $R\to R$, so in the infinite list $\text{pow}_1,\text{pow}_2,\text{pow}_3,\dots$, there must be a repeat, $\text{pow}_n=\text{pow}_m$. This means $x^n=x^m$ for all $x\in R$.

Answer 2

If $R=\mathbb{Z}/8\mathbb{Z}$, then $R^*=\{\bar1,\bar3, \bar5,\bar7\}.$ Since every element is its own inverse, we have $R^*\ncong \mathbb{Z}/4\mathbb{Z}$.

If $R=Mat_n(\mathbb{F}_q)$, then $R^*=GL_n(\mathbb{F}_{q})$, where $\mathbb{F}_{q}$ is the finite field of $q$ elements.

Therefore, the group of units of a finite ring need not be cyclic nor Abelian if the ring is noncommutative. If $R$ is a finite field, then your statement about the group of units is correct.