Question about proof of n-th orthogonal group In the book of Abstract Algebra I'm seeing there is an example of a group, $Ort_n(\mathbb{R})$ which is the group of all orthogonal endomorphisms in $\mathbb{R^n}$, all endomorphisms $f: \mathbb{R^n} \rightarrow \mathbb{R^n}$ such that $||f(u)|| = ||u||$ for all $u \in \mathbb{R^n}$ with the operation composition. When proving that this indeed is a group, in order to show that $f \in \mathbb{R^n}$ has an inverse in $Ort_n(\mathbb{R})$ the book proves that every $f$ is one-to-one, therefore it's an isomorphism, and then proves that it is surjective. * * * I don't understand why does $f$ being one-to-one imply that it is an isomorphism, and neither understand the proof of surjectivity. Any hints?

$f(u)=0$ implies that $\|f(u)\|=\|u\|=0$ $Ker f=0$ since $f$ is linear it is injective. An injective linear map between finite dimensional vector space which have the same dimension is surjective and therefore bijective.