# Orthogonal group is subgroup

How to show that Orthogonal group is subgroup of 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 since isometry is injective and also from rank nullity theorem map from V to V has range of full dimension so it is surjective. But I need 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.

• Which definition of $O(V)$ do you use? Not $O(V)=\{\,\phi\in GL(V)\mid \forall x,y\in V\colon \langle \phi x,\phi y\rangle=\langle x, y\rangle\,\}$? – Hagen von Eitzen Jun 15 '18 at 18:16
• yes b(⟨ϕx,ϕy)=b(x,y) isometry of bilinear forms – maths student Jun 15 '18 at 18:18
• Or dou you define $O(V)=\{\,\phi\in \operatorname{End}(V)\mid \forall x,y\in V\colon \langle \phi x,\phi y\rangle=\langle x, y\rangle\,\}$? In that case, you need to add the assumption that $\dim V<\infty$ to make it a group. – Hagen von Eitzen Jun 15 '18 at 18:21
• Can it be group for infinite dimensional case? – maths student Jun 15 '18 at 18:25
• I want to show it for both finite as well as infinite dimensional case. – maths student Jun 15 '18 at 18:26

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 $GL(V)$.