$\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$The compact simple Lie groups $\SO\_8(\mathbb{R}) $ and $\SO\_9(\mathbb{R}) $ both have rank 4. The group
$$
G=\SU\_3 \times \SU\_2 \times \operatorname{U}\_1 
$$
also has rank 4. Does there exist a subgroup of $\SO\_8(\mathbb{R}) $ or $\SO\_9(\mathbb{R}) $ isomorphic to $ G $?


I already know that there is an inclusion of $ G $ into $\SU\_5 $ given by putting $ \SU\_3 $ in the top 3 by 3 block, $\SU\_2 $ in the bottom 2 by 2 block and then $ U\_1 $ corresponds to the block scalar subgroup $\exp(-\frac{2 \pi i t}{3})I\_3 \times \exp(\frac{2 \pi i t }{2}) I\_2 $ where $ I\_n $ are identity matrices of the appropriate size and $ t $ parameterizes the subgroup. Since $ \SU\_5 $ naturally embeds in $\SO\_{10}(\mathbb{R}) $ we can compose embeddings and get an embedding of $ G $ into $\SO\_{10}(\mathbb{R}) $.



Smallest dimension for faithful orthogonal representation
I made some mistakes in my first version of this answer (including giving the opposite reply …), so hopefully this is correct. Thanks to @ZoltanZimboras for help sorting it out (although, of course, if it is still wrong, then the fault is entirely mine).


I claim that there is no embedding of $G$ in $\operatorname{SO}\_9(\mathbb R)$ (and hence none in $\operatorname{SO}\_8(\mathbb R)$).


If $\{\alpha\_1, \alpha\_2, \alpha\_3, \alpha\_4\}$ is a system of simple roots for $\mathsf B\_4$, with $\alpha\_4$ short, then the subsystems generated by $\{\alpha\_1, \alpha\_2\}$ and $\{\alpha\_4\}$ are orthogonal and of types $\mathsf A\_2$ and $\mathsf A\_1$, respectively. I claimed that this gave an embedding of $\operatorname{SU}\_3 \times \operatorname{SU}\_2$ in $\operatorname{SO}\_9(\mathbb R)$, but @ZoltanZimboras pointed out that it actually only gives an embedding in $\operatorname{Spin}\_9(\mathbb R)$ (or whatever is the correct notation for the compact spin group). Note that, under this embedding, the image $\alpha\_4^\vee(-1)$ of $-1 \in \operatorname{SU}\_2$ in $\operatorname{Spin}\_9(\mathbb R)$ lies in the kernel of $\operatorname{Spin}\_9(\mathbb R) \to \operatorname{SO}\_9(\mathbb R)$, so we cannot simply compose with the projection to get an embedding in $\operatorname{SO}\_9(\mathbb R)$. The additional $\operatorname U\_1$ can be taken to be the image of $2\alpha\_1^\vee + 4\alpha\_2^\vee + 6\alpha\_3^\vee + 3\alpha\_4^\vee$.


We do not have an embedding of $G$ in $\operatorname{Spin}\_8(\mathbb R)$, since the absolute root system of $\operatorname{Spin}\_8(\mathbb R)$ is of type $\mathsf D\_4$, but $\mathsf D\_4$ contains no subsystem of type $\mathsf A\_2 + \mathsf A\_1$. (The orthogonal complement of a root in $\mathsf D\_4$ is of type $\mathsf A\_1 + \mathsf A\_1$.)


Suppose we had an embedding $G \to \operatorname{SO}\_9(\mathbb R)$. We claim that some absolute root of $G$ is short in the absolute root system of $\operatorname{SO}\_9(\mathbb R)$. Indeed, suppose not, and consider one of the roots of the $\operatorname{SU}\_2$ factor. Since it is long, its orthogonal complement in $\mathsf B\_4$ is of type $\mathsf B\_2 + \mathsf A\_1$, in Carter's notation (so the $\mathsf A\_1$ roots are long). The long roots in $\mathsf B\_2 + \mathsf A\_1$ form a system of type $3\mathsf A\_1$, which contains no subsystem of type $\mathsf A\_2$.


Let $\alpha$ be an absolute root of $G$ that is short in $\operatorname{SO}\_9(\mathbb R)$. Then the simple subgroup of $\operatorname{SO}\_9(\mathbb R)$ whose absolute root system is $\{\pm\alpha\}$ is adjoint (i.e., $\operatorname{SU}\_2/\langle-1\rangle$), but the simple subgroup of $G$ whose absolute root system is $\{\pm\alpha\}$ is simply connected (i.e., $\operatorname{SU}\_2$). (See [@nfdc's answer](https://mathoverflow.net/a/270280) to [Centralizers of subtori in reductive groups, derived subgroups](https://mathoverflow.net/questions/270205/centralizers-of-subtori-in-reductive-groups-derived-subgroups) .) Since these are the same group, we have a contradiction.


@ZoltanZimboras also asked, in a comment on the first version of the answer, how the resulting representations $\operatorname{SU}\_3 \to \operatorname{Spin}\_9(\mathbb R) \to \operatorname{SO}\_9(\mathbb R)$ and $\operatorname{SU}\_2 \to \operatorname{Spin}\_9(\mathbb R) \to \operatorname{SO}\_9(\mathbb R)$ decompose. I couldn't answer most questions about this representation, but, fortunately, since all computations have been in terms of what happens on tori, I can answer this one! We have that $(\alpha\_1^\vee, \alpha\_2^\vee) : (\operatorname S^1)^2 \to \operatorname{Spin}\_9(\mathbb R)$ and $\alpha\_4^\vee : \operatorname S^1 \to \operatorname{Spin}\_9(\mathbb R)$ factor through embeddings of maximal tori $(\operatorname S^1)^2 \to \operatorname{SU}\_3$ and $\operatorname S^1 \to \operatorname{SU}\_2$, which I will use as coördinates for those tori.


The weights of the (implicit) chosen maximal torus in $\operatorname{Spin}\_9(\mathbb R)$, in the natural representation *via* its projection to $\operatorname{SO}\_9(\mathbb R)$, are the trivial weight, together with $\alpha\_i + \dotsb + \alpha\_4$ and their opposites for $i \in \{1, 2, 3, 4\}$. Thus the weights of $\operatorname{SU}\_3 \times \operatorname{SU}\_2 \times \operatorname U\_1$ are the trivial weight $(0, 0; 0; 0)$, together with $(1, 0; 0; 2)$, $(-1, 1; 0; 2)$, $(0, -1; 0; 2)$, $(0, 0; 2; 0)$, and their opposites. Thus, if I haven't mixed up my computation, the restricted representation of $G$ is the sum of the trivial representation; the two fundamental representations of $\operatorname{SU}\_3$, on which $\operatorname{SU}\_2$ acts trivially, and on one of which the weight of $\operatorname U\_1$ is $2$ and on the other of which it is $-2$; and the representation of $\operatorname{SU}\_2$ with highest weight $2$, on which $\operatorname{SU}\_3 \times \operatorname U\_1$ acts trivially.


