It is east to use the special unitary group to contain the special orthogonal group so $$SU(n) \supset SO(n) .$$ For $n=8$, we have $$SU(8) \supset SO(8).$$
We know that the $SO(8)$ has a double cover $Spin(8)$ so $Spin(8)/(\mathbf{Z}/2)=SO(8)$ and also there is a $PSO(8)=SO(8)/(\mathbf{Z}/2)$.
My question is that does $$SU(8) \supset Spin(8)?$$ $$SU(8) \supset SO(8)/(\mathbf{Z}/2)?$$
What are some relations that we can say between $SU(8)$, and $Spin(8)$ and $SO(8)/(\mathbf{Z}/2)$?
The compact group $\mathbf{Spin}(8)$ cannot be embedded in $\mathbf{SU}(8)$, nor can its complexification $\mathbf{Spin}(8,\Bbb C)$ be embedded in $\mathbf{SL}(8,\Bbb C)$ (the two questions are equivalent, via the correspondence between complex groups and their maximal compact subgroups).
To see why, consider the second (complex group) formulation; for an embedding into $\mathbf{SL}(8,\Bbb C)$ would a fortiori be one into $\mathbf{GL}(8,\Bbb C)$, in other words a faithful (complex) representation of $G=\mathbf{Spin}(8,\Bbb C)$ of dimension$~8$.
A first important point is the fact that the centre of $G$ is a Klein $4$-group (in other words isomorphic to $(\Bbb Z/2\Bbb Z)^2$). Indeed the centre of $\mathbf{SO}(8,\Bbb C)$ has two elements, the nontrivial one being $-I$ (since $8$ is even), and the two-fold cover of $\mathbf{SO}(8,\Bbb C)$ by $G$ gives another central element of order$~2$, by which every central element of $\mathbf{SO}(8,\Bbb C)$ gives rise to two central elements of $G$. It can be verified that these four elements form a Klein $4$-group rather than a cyclic group (as happens for spin groups of the form $\mathbf{Spin}(4n+2,\Bbb C)$).
Now the dual group of the centre of $G$ (non canonically isomorphic to it) can be realised as the quotient of the character lattice $X^*$ of by the root lattice$~R$. Now the fact that this is not a cyclic group means that no irreducible representation of$~G$ can be faithful: since all weights of an irreducible representation lie in the same $R$-coset, the sub-lattice of $X^*$ that they span projects to a cyclic subgroup of $X^*/R$, which cannot be all of that quotient, which means there is some nontrivial central element of $G$ on which all those weights vanish, so that it acts trivially in the representation. It follows that although three of the $4$ fundamental representations of$~G$ have dimension$~8$ (the "$\mathbf{SO}(8)$" representation, which is the first fundamental one, and the last two fundamental representations), none of them are faithful. Indeed, the smallest faithful representations of$~G$ are reducible $16$-dimensional representations that are sums of two distinct fundamental ones; the classical (Clifford algebra) realisation of $G$ being obtained from the sum of the two last fundamental representations.
Similar things can be said for even spin groups of higher rank, though the dimensions grow wider apart: while the first fundamental representation of $\mathbf{Spin}(2n,\Bbb C)$ has dimension$~2n$, giving rise to the double covering map $\mathbf{Spin}(2n,\Bbb C)\to\mathbf{SO}(2n,\Bbb C)$, the smallest faithful representation is either (1) if $n$ is odd: of dimension $~2^{n-1}$ being one of the last two fundamental (called half-spin) representations, or (2) if $n$ is even: of dimension$~2^n$, being the sum of the last two fundamental representations (since for $n$ even the half-spin representations are not faithful).
To complete the story for $\mathbf{PSO}(8,\Bbb C)$, which is the quotient of $\mathbf{SO}(8,\Bbb C)$ by its centre$~\{I,-I\}$, the lowest dimensional non trivial representation of this group is the adjoint representation, which is of dimension$~28$ (so in particular $\mathbf{PSO}(8,\Bbb C)$ does not embed into $\mathbf{SL}(8,\Bbb C)$). To see that this is the case, the representations of $\mathbf{PSO}(8,\Bbb C)$ lift to representations of $\mathbf{Spin}(8,\Bbb C)$ for which the $4$ elements of the centre of the latter act as identity. This means that all weights of the representation lie in the root lattice, which is the sub-lattice of weights for $\mathbf{Spin}(8,\Bbb C)$ that (seen as group morphisms from its maximal torus to$~\Bbb C^\times$) take the value$~1$ on the whole centre. Every irreducible representation has a highest weight, which is a non-negative integer combination of fundamental weights, and the dimension increases monotonically with the coefficients of this combination, so we can look at small non negative coefficients (for the zero highest weight we have the trivial representation, which we will ignore). As mentioned above, three of the four fundamental weights correspond (as highest weights) to representations of dimension$~8$, but in none of these the whole centre acts trivially: indeed in each case the subgroup acting trivially is a different $2$-element subgroup of the Klein $4$-group centre. The remaining fundamental weight (the one for to the "branch node" of the $D_4$ Dynkin diagram) does lie in the root lattice; indeed the corresponding representation is the adjoint representation of $\mathbf{Spin}(8,\Bbb C)$, whose dimension is$~28$ ($4$ for the maximal torus plus $24$ for the roots). Therefore in order to have trivial action of the centre, the coefficients of a highest weight of an irreducible representation must be even for each of the $3$ "leaf" fundamental weights. The only combination that could compete with the "branch" fundamental weight then is twice a "leaf" fundamental weight, but this turns out to give, in each of the three cases, a representation of dimension$~35$, so this is not better.
