Does the embedding of $M_{n}(\mathbb C)$ into $M_{2n}(\mathbb R)$ send $GL_n(\mathbb C)$ into $GL_{2n}(\mathbb R)$? We can embed $M_{n}(\mathbb C)$ into $M_{2n}(\mathbb R)$ via
$$Z=X+iY \mapsto \pmatrix{X& Y \cr -Y& X}, $$
where $X,Y \in M_{n}(\mathbb R)$.  It can be seen that this embedding is an injective ring homomorphism that sends the multiplicative identity of $M_n(\mathbb{C})$ to the multiplicative identity of $M_{2n}(\mathbb{R})$.

Does this embedding also send $\text{GL}_{n}(\mathbb C)$ into $\text{GL}_{2n}(\mathbb R)$?
Namely how do I conclude from
$$\det (X+iY)\ne 0$$ that $$\det \pmatrix{X& Y \cr -Y& X}\ne 0\,?$$

 A: This is immediate from the fact that your embedding is a ring homomorphism.  If $X+iY$ is invertible with inverse $U+iV$, then $\begin{pmatrix}U & V \\ -V & U\end{pmatrix}$ is inverse to $\begin{pmatrix}X & Y \\ -Y & X\end{pmatrix}$, and in particular $\begin{pmatrix}X & Y \\ -Y & X\end{pmatrix}$ is invertible.
A: Let $I$ be the identity matrix with the same dimension as that of $X$ and $Y$.  Note also that
$$\begin{bmatrix}\vphantom{\frac{1}{\sqrt{2}}}X&Y\\-Y&\vphantom{\frac{1}{\sqrt{2}}}X\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}}I&-\frac{\text{i}}{\sqrt{2}}I
\\
-\frac{\text{i}}{\sqrt{2}}I&\frac{1}{\sqrt{2}}I\end{bmatrix}\,\begin{bmatrix}\vphantom{\frac{1}{\sqrt{2}}}X-Y\text{i}&0\\0&\vphantom{\frac{1}{\sqrt{2}}}X+Y\text{i}\end{bmatrix}\,\begin{bmatrix}\frac{1}{\sqrt{2}}I&+\frac{\text{i}}{\sqrt{2}}I
\\
+\frac{\text{i}}{\sqrt{2}}I&\frac{1}{\sqrt{2}}I\end{bmatrix}\,.$$
The matrix $J:=\begin{bmatrix}\frac{1}{\sqrt{2}}I&-\frac{\text{i}}{\sqrt{2}}I
\\
-\frac{\text{i}}{\sqrt{2}}I&\frac{1}{\sqrt{2}}I\end{bmatrix}$ has determinant $1$ because
$$\begin{align}\det(J)&=\det\begin{bmatrix}\frac{1}{\sqrt{2}}I&-\frac{\text{i}}{\sqrt{2}}I
\\
0&\frac{1}{\sqrt{2}}I+\frac{1}{\sqrt{2}}I\end{bmatrix}\quad\text{(adding $\text{i}$ times the first row to the second row})
\\&=\det\begin{bmatrix}\frac{1}{\sqrt{2}}I&-\frac{\text{i}}{\sqrt{2}}I
\\
0&\vphantom{\frac{1}{\sqrt{2}}}\sqrt{2}\,I\end{bmatrix}\\&=\det\left(\frac{1}{\sqrt{2}}I\right)\cdot\det(\sqrt{2}\,I)=\det\left(\frac{1}{\sqrt2}\,I\cdot \sqrt{2}\,I\right)=\det(I)=1\,.\end{align}$$
Thus, $\bar{J}=\begin{bmatrix}\frac{1}{\sqrt{2}}I&+\frac{\text{i}}{\sqrt{2}}I
\\
+\frac{\text{i}}{\sqrt{2}}I&\frac{1}{\sqrt{2}}I\end{bmatrix}$ has determinant $$\det(\bar{J})=\overline{\det(J)}=\bar{1}=1\,.$$
Hence,
$$\begin{align}\det\begin{bmatrix}X&Y\\-Y&X\end{bmatrix} &= \det(J)\det(X-Y\text{i})\det(X+Y\text{i})\det(\bar{J})\\&=1\cdot\overline{\det(X+Y\text{i})}\cdot\det(X+Y\text{i})\cdot1=\Big|\det(X+Y\text{i})\Big|^2\,.\end{align}$$
