Interchanging field axioms on finite set

Question

Let $F$ be a FINITE set with two operations "+" and "$\cdot$" satisfying the field axioms but the requirement, for every $a \in F\setminus\{0\}$ exists an element $b\in F$ such that $b\cdot a =1$, where $1$ is the multiplicative identity, is interchanged with the requierement, that for all $a,b \in F\setminus \{0\}$ we have $ab\in F \setminus \{0\}$.

We have to show, that $F$ is still a field.

My idea was to show that there exist $n,m \in \mathbb{N}$ such that $a^n=a^m$ and thus $a^n=a^{m-n}a^n$. But I don't if this implies $a^{m-n}=1$ and thus $a^{-1}=a^{m-n-1}$.

Answer 1

For $a\neq0$, the map on $F-\{0\}$, $b\mapsto ba$, is injective ($ba=ca$ implies $(b-c)a=0$, so by assumption $b=c$). Thus it is surjective, since $F$ is finite. Hence there exists $b\neq0$ with $ab=1$.

Answer 2

A commutative ring $F$ with the properties $0 \neq 1$ and $\forall a,b \in F\setminus \{0\}:ab \in F\setminus\{0\}$ is called an integral domain. So the task at hand is to show that a finite integral domain is actually a field.

Your idea works if we use an extra trick: for $a \neq 0$ the pigeonhole principle gives $m, n \in \Bbb N$ with $n <m$ such that $a^n =a^m=a^n a^{m-n}$, so $a^n(a^{m-n}-1)=0$. Since a product of nonzero elements is nonzero and $a \neq 0$, $a^n \neq 0$, so the only way the product with $a^{m-n}-1$ can be zero is if $a^{m-n}-1=0$, so $a^{m-n-1}a=1$ which shows that $a$ is invertible.