# Prove SU(n) is a subgroup of SO(2n)

I want to show that SU(n) is a subgroup of SO(2n). I can show the first step, that is, for any $$U=U_R + iU_I\in SU(n)$$ with $$U_R,U_I\in\mathbb{R}^{n\times n}$$, we identify $$U$$ with the real $$2n\times 2n$$ matrix $$X = \begin{pmatrix} U_R & - U_I\\ U_I & U_R \end{pmatrix}$$ and I can show that $$X$$ is orthogonal, i.e., $$XX^T= Id_{2n\times 2n}$$. But I do not know how to show that $$\det X = 1$$. Can someone show we how?

• I think that we always have $\det X=|\det U|^2$. For all $U\in GL_n(\Bbb{C})$. Jul 18, 2022 at 10:10
• @JyrkiLahtonen more generally, if $L/K$ is a finite Galois extension, then by restriction of scalars any $U \in M_{n \times n}$ gives rise to a square matrix of dimension $n[L:K]$ with determinant $N_{L/K}(\det U)$ Jul 18, 2022 at 10:39
Here is one nice approach. Denote $$P = \pmatrix{I & -iI\\-iI & I},$$ note that $$P^{-1} = \frac 12 P^*$$ (where $$P^*$$ denotes the conjugate-transpose of $$P$$). Now, note that $$\det(X) = \det(P^{-1}XP)$$, and verify (via block-matrix multiplication) that $$P^{-1}XP = \frac 12 P^*XP = \pmatrix{U_R + iU_I & 0\\0 & U_R - iU_I}.$$ Thus, we have $$\det(X) = \det(U) \det(\bar U) = \det(U)\overline{\det(U)} = |\det(U)|^2 = 1.$$ (In the above, $$\bar \cdot$$ denotes the complex conjugate, so $$\bar U = U_R - iU_I$$).