How to show that Orthogonal group is a subgroup of the general linear group GL(V), where V is vector space?
Since general linear group GL(V) group of all linear transformation which are bijective i.e Invertible.
I know that an isometry is injective and also from the rank-nullity theorem that the image under the map from V to V has the dimension of the full group so it is surjective. But I need a more formal proof.
Also since to show it is subgroup we have show x*inverse of y belongs to Orthogonal group O(V) by using bilinear form isometry. Please help me out to show it is surjective using bilinear forms and show it is subgroup.
Let $\phi\in O(V)=\{\,\phi\in GL(V)\mid \forall x,y\in V\colon \langle \phi x,\phi y\rangle=\langle x, y\rangle\,\}$. Then in particular $\phi\in GL(V)$ and $\phi^{-1}$ exists. Then for all $x,y\in V$, $$\langle \phi^{-1} x,\phi^{-1} y\rangle=\langle \phi\phi^{-1} x,\phi\phi^{-1} y\rangle=\langle x,y\rangle$$ and we conclude $\phi^{-1}\in O(V)$. If also $\psi\in O(V)$, then for all $x,y\in V$, $$ \langle\psi\phi x,\psi\phi y\rangle =\langle \phi x,\phi y\rangle =\langle x,y\rangle$$ and we conclude $\psi\circ \phi\in O(V)$. Trivially, the identity is $\in O(V)$. So $O(V)$ is a non-empty subset of the group $GL(V)$ and closed under composition and taking inverses. We conclude that $O(V)$ is a subgroup of $GL(V)$.