Let $H(\mathbb{R})=\{a+bi+cj+dk \mid i^2=j^2=k^2=-1, ij=-ji=k, a, b, c,d \in \mathbb{R}\}$. Let $ZH(\mathbb{R})$ be the center of $H(\mathbb{R})$.
How to show that $H(\mathbb{R})/ZH(\mathbb{R}) \cong SO(3, \mathbb{R})$?
I think that $ZH(\mathbb{R})=\mathbb{R}$, $$ SO(3, \mathbb{R}) = \{g \in GL(3, \mathbb{R}) \mid gg^T=I, \det(g)=1\}. $$ An element of $H(\mathbb{R})/ZH(\mathbb{R})$ is of the form $bi+cj+dk$. What is the corresponding element of $bi+cj+dk$ in $SO(3, \mathbb{R})$? Thank you very much.
First, I believe you mean to consider to the Lie group $G=H(\mathbb{R}-\{0\})$, under multiplication.
Consider the vector space $V^3 = \operatorname{span}_{\mathbb{R}}\{i,j,k\}$ where we consider $i$, $j$, and $k$ an orthonormal basis.
Define a map $\phi:G\rightarrow SO(V)$ by $\phi(q)(v) = qvq^{-1}$.
I claim that $\phi$ is a surjective homomorphism with $\ker \phi = Z(G)$. If we can prove this, this implies that $G/Z(G)\cong SO(V)$.
To begin with, note that $|\phi(q)(v)| = |qvq^{-1}| = |q||v||q^{-1}| = |v|$, so $\phi(q)$ really is an isometry. In particular, $\phi(q)\in SO(V)$.
Now we show $\phi$ is a homomorphism. We have for any $q_1,q_2\in G$ and $v\in V$, \begin{align*} \phi(q_1 q_2)(v) &= (q_1 q_2) v (q_1 q_2)^{-1}\\ &= q_1 q_2 v q_2^{-1}q_1^{-1}\\ &= q_1(\phi(q_2)(v)) \\ &= \phi(q_1)\phi(q_2)(v). \end{align*}
Now we show the kernel of $\phi$ is $Z(G)$. First, if $q\in Z(G)$, then $\phi(q)(v) = qvq^{-1} = qq^{-1} v = v$, so $\phi(q) = I$, that is, $q\in\ker \phi$. Conversely, if $q$ is not in $Z(G)$, then there is an $r$ for which $qr\neq rq$. By inspecting real and imaginary party separately, it follows that $q\operatorname{Im}(r) \neq \operatorname{Im}(r)q$. Thus, $\phi(q)(\operatorname{Im}(r))\neq r$, so $\phi(q)\neq I$, so $\phi(q)$ is not in the kernel of $\phi$.
All we have left is surjectivity. This ends up being a bit more delicate. Probably the easiest way to prove it's surjective is to note that since $\phi$ is a homomorphism, $\phi(G)$ is a subgroup of $SO(V)$ and that, further, there is a small neighborhood around $I\in SO(V)$ contained in $SO(V)$. (The second fact can be proven by understanding the induced maps at the Lie algebra level.) Then, one proves via an open-closed argument that the group generated by any open set containing the identity element in fact contains the entire identity component. This last fact hold for any Lie group.