Prove $SU(n)\times Spin(k) \subset \frac{{Spin}(2n)\times Spin(k)}{\mathbb{Z}/2}$ for sufficient $n$ and $k$? We knew that 


*

*$SU(n) \subset {Spin}(2n)$ is true from Is $SU(n) \subset \text{Spin}(2n)$?
also


*

*$SU(n) \subset {SO}(2n)= \frac{{Spin}(2n)}{\mathbb{Z}/2}$ is true from $U(n)$ is a subgroup of $SO(2n)$
I would like to generalize the result.
Question: I hope to prove or disprove that

$$
SU(n)\times Spin(k) \subset \frac{{Spin}(2n)\times Spin(k)}{\mathbb{Z}/2}?
$$
Does this hold for general $n$ or $k$? or some range of $n\geq 2,k\geq 3$?

here $\mathbb{Z}/2$ is the order 2 normal subgroup of ${Spin}(2n)$ also the order 2 normal subgroup of $ Spin(k)$. So we mod out the common $\mathbb{Z}/2$. Namely, we consider a more exotic $\frac{{Spin}(2n)\times Spin(k)}{\mathbb{Z}/2}\neq  {SO}(2n)\times Spin(k)$ and $\frac{{Spin}(2n)\times Spin(k)}{\mathbb{Z}/2}\neq  {Spin}(2n)\times SO(k)$, we do not consider the product of two gruops.
I check that for lower dimensions $n\le 6$ there are accidental isomorphisms of spin groups with other familiar Lie groups: $\DeclareMathOperator{Spin}{\mathrm{Spin}}$
$$\begin{array}{|l|l|} \hline Spin(1) & {O}(1)=\mathbb{Z}/2 \\ \hline Spin(2) & {SO}(2)=U(1) \\ \hline Spin(3) & {Sp}(1)=SU(2) \\ \hline Spin(4) & {Sp}(1)\times{Sp}(1)=SU(2) \times SU(2) \\ \hline Spin(5) & {Sp}(2) \\ \hline Spin(6) & {SU}(4) \\ \hline \end{array} $$
Here I list down the Lie group isomorphisms, not the Lie algebra.
So we can check:
$n=2,k=3$,
$$
SU(2)\times Spin(3) \subset \frac{{Spin}(4)\times Spin(3)}{\mathbb{Z}/2}=
\frac{SU(2)\times SU(2)\times Spin(3)}{\mathbb{Z}/2} \text{ is true}?
$$
How about general $n\geq 2,k\geq 3$?
 A: First, the center of $Spin(4k+2)$ is a $\mathbb{Z}_4$, and the center of $Spin(odd)$ is a $\mathbb{Z}_2$, so there is a unique $\mathbb{Z}_2$ you can quotient by in each of these cases.  But for $Spin(4k)$, the center is $\mathbb{Z}_2\oplus \mathbb{Z}_2$, so there is ambiguity when you say "the order two normal subgroup of $Spin(4k)$".  I'll assume you're referring to the $\mathbb{Z}_2$ for which $Spin(4n+2)/\mathbb{Z}_2 \cong SO(4n+2)$.
With that out of the way, we claim that:

For any $n\geq 2$ and $k\geq 3$, there is a Lie subgroup of $G:= (Spin(2n)\times Spin(k))/\mathbb{Z}_2$ which is isomorphic to $H:=SU(n)\times Spin(k)$.

Before proving this, recall that in your previous question, I showed that there is an $SU(n)$ in $Spin(2n)$ obtained by lifting the $SU(n)$ in $SO(2n)$.  It follows that if $p:Spin(2n)\rightarrow SO(2n)$ is the double cover map, then $p|_{SU(n)}$ is injective.
Now, let's prove the theorem.
Proof.  Let $\overline{G} = Spin(2n)\times Spin(k)$ be the universal cover of $G$, and suppose $\pi:\overline{G}\rightarrow G$ is the double covering map with kernel given by $\mathbb{Z}_2 = \langle z\rangle$.
In your previous question, we already established that there is a subgroup of $Spin(2n)$ isomorphic to $SU(n)$.  Now, let $\overline{H} = SU(n)\times Spin(k)\subseteq Spin(2n)\times Spin(k)$.  Obviously, $\overline{H}$ is isomorphic to $H$.
Consider $\pi(\overline{H})\subseteq G$.  We claim that $\pi(\overline{H})$ is isomorphic to $H$.  To that end, note that $\pi|_{\overline{H}}:\overline{H}\rightarrow \pi(\overline{H})$ is a surjective Lie homomorphism by definition, so we just need to check that it is injective, and then we'll have the isomorphisms $H\cong \overline{H}\cong \pi(\overline{H})$.
Since $\ker \pi_{\overline{H}}\subseteq \ker \pi$, we need only show $z\notin \ker \pi|_{\overline{H}}$.  So, assume for a contradiction that $z\in \ker \pi|_{\overline{H}}\subseteq \overline{H}$.  Letting $\rho:\overline{G}\rightarrow Spin(2n)$ be the projection map, we see that $\rho(z)\in \rho(\overline{H}) = SU(n)\subseteq Spin(2n)$.
But the projection map $p:Spin(2n)\rightarrow SO(2n)$ is simply $Spin(2n)\rightarrow Spin(2n)/\langle\rho(z)\rangle  \cong SO(2n)$.  If $\rho(z)\in SU(n)\subseteq Spin(2n)$, then the projection map $p$ restricted to $SU(n)$ is not injective, contradicting the fact mentioned before the proof.  This establishes that $z\notin \ker\pi_{\overline{H}}$, so $\pi_{\overline{H}}$ is a Lie isomorphism from $\overline{H}$ to $\pi(\overline{H})$.  $\square$.
