Properties of finite ring

Question

Suppose $R$ is a finite ring. It may be commutative or it may be not. Let $x \in R$.

1. Show that there exists a positive integer $n$ such that $x^n = x^{2n}$.
2. Is it true that $x^{k!} = x^{2k!}$, where $k = |R|$?

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In the first part should we take random $x$ and consider its powers, then use pigeonhole principle, to show that some powers must coincide? Or is there better way to tackle first problem?

For the second problem do we have to provide a counter-example in form of the ring modulo prime and show that if first statement is true second will not hold?

Answer 1

1. Let $x\in R$. Since $R$ is finite, there are some positive integers $k$ and $l$ such that $x^k=x^{k+l}$. Then $$(x^{kl})^2=x^{kl}x^{kl}=x^{k(l-1)}(x^k x^l)x^{l(k-1)}=x^{k(l-1)}x^{k+l}x^{l(k-1)}=x^{k(l-1)}x^{k}x^{l(k-1)}=x^{kl}x^{l(k-1)}=\cdots =x^{kl},$$ so $x^{kl}$ is an idempotent.

2. Set $n=|R|$. For $x\ne 0$ we can choose $k,k+l\in\{1,\dots,n\}$ since the set $\{0,x,x^2,\dots,x^n\}$ has $n+1$ elements. (If $x^m=0$ for some $1<m\le n$, then $x^m=x^{2m}$.) Then $kl\mid n!$. This shows that $x^{n!}=x^{2n!}$ (for all $x\in R$).

Answer 2

Building on @user26857's answer: it is possible to find $k, \ell$ such that $k + \ell \leq |R|$. Intuitively, $\ell$ is the length of the cycle containing $x^k$, which cannot exceed $|R|$, and $k$ is the distance from $x$ to an element that is part of a multiplicative cycle; by the pigeonhole principle we can assume minimal values such that $k \leq |R| - \ell$.

If $k \neq \ell$, then $k \ell$ divides $|R|!$, hence for any natural number $z$, $x^{z|R|!}=x^{zm(k\ell)}=x^{k\ell}$.
If $k = \ell$, then $k, \ell \leq |R| / 2$, and $k$ and $2\ell$ are both terms in the product $|R|!=(|R|)(|R|-1)\cdots(1)$; again, $k\ell$ divides $|R|!$, and we have $x^{z|R|!}=x^{zm^\prime(k\ell)}=x^{k\ell}$.
In either case, $x^{|R|!}=x^{2|R|!}$.