Question: $\DeclareMathOperator{\Spin}{Spin}$
Does $SU(n) \to \Spin(2n)$ induces an embedding of $SU(n) \subset \Spin(2n)$?
Do $SU(n) \to O(2n)$ induces an embedding of $SU(n) \subset O(2n)$?
Some info:
Here the spin group $\Spin(2n)$ has a $\mathbf{Z}_2$ center as a normal subgroup such that $\Spin(2n)/\mathbf{Z}_2=SO(2n).$ Potentially there is a lift as a commutative diagram. $$ \begin{array}{ccc} & & \Spin(2n)\\ &\nearrow & \downarrow\\ SU(n) & \longrightarrow & SO(2n) \end{array}. $$
Here the orthogonal group $O(2n)$ has a normal $SO(2n)$ subgroup such that $O(2n)/SO(2n)=\mathbf{Z}_2.$ Potentially there is a lift (not sure my description of lift is precise?) as a commutative diagram. $$ \begin{array}{ccc} & & O(2n)\\ &\nearrow & \uparrow\\ SU(n) & \longrightarrow & SO(2n) \end{array}. $$
We can take the complex vector space $\mathbb C^n$ as $\mathbb R^{2n}$ and recall that the standard inner product on $\mathbb R^{2n}$ is the real part of the standard Hermitian inner product on $\mathbb C^n$. So this shows that $$SU(n) \subset SO(2n)$$ (maybe this also proves that $U(n) \subset O(2n)$ as well? please clarify.)
I know the first homotopy group that $$\pi_1(SU(n))=0, \; \forall n \geq 1$$ $$\pi_1(SO(2))=\pi_1(O(2))=\pi_1(\Spin(2))=\mathbf{Z}, \; $$ $$\pi_1(SO(m))=\pi_1(O(m))=\mathbf{Z}_2, \; \forall m \geq 3$$ $$\pi_1(\Spin(m))=0, \; \forall m \geq 3.$$ Perhaps the fundamental group is crucial to understand whether the lift and the embedding are possible.