I am wanting to
Show that a map $f' : S^3\times S^3\rightarrow SO(4)$ induced from a map $f : S^3\times S^3 \rightarrow GL_4(\mathbb{R}) $ is a homomorphism where $S^3$ is a Lie group of unit quaternion.
Here, the map $f$ is defined by
$$f(a,b)\cdot x = axb^{-1} $$ where $x\in \mathbb{H}=\mathbb{R}^4$.
I have prove that $f$ is a homomorphism since
$$f(aa',bb')\cdot x = (aa')x(bb')^{-1}\\=a(a'xb'^{-1})b^{-1}=f(a,b)\cdot (a'xb'^{-1})=f(a,b)\cdot (f(a',b')\cdot x)=f(a,b)f(a',b')\cdot x .$$
I know that each $f(a,b)$ is invertible, so $f(a,b)\in GL_4(\mathbb{R})$ but I cannot see why it should be in $SO(4)$. Is there any way to see this?
Any help would be very thankful!
Note that $f(a,b)\in GL^+$ because $f$ is continuous, $S^3$ is connected, and $f(1,1)$ is the identity. So it suffices to prove that $f(a,b)$ is an isometry. This follows simply from $\mathbb{H}$ being a normed division algebra: $$\lvert f(a,b)(x)\rvert = \lvert axb^{-1}\rvert = \lvert a\rvert\cdot\lvert x\rvert\cdot\lvert b^{-1}\rvert = \lvert x\rvert$$