Here is the question that I want to answer part(c) in it:
Define $E \in GL_{2}(\mathbb{R})$ by $E = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and let $\mathcal{R} = \{aI + bE| a,b \in \mathbb{R}\} \subset M_{2}(\mathbb{R}).$
$(a)$ Show that $\mathcal{R} \cong \mathbb{C}$ as rings (so $\mathcal{R}$ is a field). Which matrices correspond to the subgroup $S^{1} \subset \mathbb{C}^{*}$?
$(b)$ Let $\mathbb{H} \subset M_{2}(\mathbb{C})$ be the set of matrices of the form: $$ \begin{pmatrix} z & - \bar{\omega} \\ \omega & \bar{z} \end{pmatrix} \quad \quad z, \omega \in \mathbb{C}$$Show that $\mathbb{H}$ is a division ring. ( $\mathbb{H}$ is called the \textbf{quaternion algebra}).
$(c)$ Find a way to represent $\mathbb{H}$ as a subring of $M_{4}(\mathbb{R}).$\ (Hint: Combine parts $(a)$ and $(b)$)
My question is:
If I am fine with proving $(a)$ and $(b),$ how can I combine parts $(a)$ and $(b)$ to answer $(c)$? I got the following hint **Consider 2 elements $A$ and $B^{-1}$ belong to $H$ if their product $AB^{-1}$ also belong to $\mathbb{H}$ then \mathbb{H} is a subring of $M_{2}(C)$ if field is $\mathbb{R}$ then it would be a subring of $M_{4}(\mathbb{R})$, ** but still I do not know how to show it, could anyone help me please?
Here is more general approach:
Let $F$ be a field and let $A$ be a finite dimensional algebra over $F$ of dimension $n$.
Then $\mu: A \to End_F(A)$ given by $\mu(a)(x)=ax$ is an injective ring homomorphism.
Choosing a basis for $A$ over $F$ gives a ring isomorphism $\phi:End_F(A) \cong M_n(F)$.
Thus we get an embedding $\phi \circ \mu : A \to M_n(F)$.
Apply this to $F=\mathbb R$, $A=\mathbb H$, and $n=4$. A basis for $\mathbb H$ over $\mathbb R$ is given by $\{1,i,j,k\}$ where $$ 1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad i = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad j = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad k = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, $$