Property of finite division ring

Question

If D is a finite division ring containing $q$ elements, then $a^q=a, > \forall a \in D$ ?

I wish to know is that a valid claim?

All i know is $D$ must be a finite field here, as it's a finite division ring. Also I know that every finite field has order equal to power of some prime.

I could only think of $\mathbb Z_p$ which has order $p$, and thus I applied the Fermat's Little Theorem to conclude that the claim holds true in this case. But being a layman, I am really not sure if I am correct and also I am clueless about the case when the order of the field is of the form $p^k$ where $k>1$.Some help would be really appreciated. Thanks in advance.

Answer 1

Yes. Consider the group of units $D^*$ which is a group of order $q-1$. For an element $a \in D^*=D\setminus \{0\}$, we have

$$1=a^{|D^*|}= a^{q-1}\implies a=a^q$$

and for $a=0$ the statement is trivial.

Answer 2

Well, consider the multiplicative group $F_q^*$ of a finite field $F_q$ with $q$ elements. Since it has $q-1$ elements, the theorem of Lagrange says that for each element $a\in F_q^*$, $a^{q-1}=1$. Then $a^q=a$. But this equality also holds for the zero element. Thus $a^q=a$ for each element $a\in F_q$.

Note that for each finite abelian group $G$ with $n$ elements, $g^n=1$ for each $g\in G$. This can be proven more directly using the fact that $G$ is abelian. No need for Lagrange here.