If $F$ is a field, show the following function is a permutation

Question

Let $F$ be a field. Show that the function $a\rightarrow a^{-1}$ is a permutation of $F\{0_F\}$

So I know that if it is indeed a permutation, then it is one-to-one and onto.

Also, For every $a$,$b$ element of $F\{0\}$, $a$ cannot equal $b$, and $a^{-1}$ cannot equal $b^{-1}$

So I think I need to assume that for every $b$ element of $F\{0\}$ there exists $a$ element of $F\{0\}$ s.t. $a^{-1}=b$. Any insight on how to continue?

Answer 1

The function $ a\mapsto a^{-1}$ is its own inverse and so is a bijection.

Answer 2

By using defination of field you can say that $F-\{0 \}$ is a group under multiplication, after that you can conclude $x\mapsto x^{-1}$ is an bijection as every element has uniqe inverse in a group.