How do we embed a Spin group to a Unitary group?
Say a Spin(10) group is a Lie group with $\frac{10 \cdot 9}{2}=45$ Lie algebra generators.
Say a special unitary group SU($n$) has a $n^2-1$ Lie algebra generators.
When $n \geq 7$, we have $$ n^2-1 \geq 7^2-1=48>45 $$
questions:
What is the minimal $n$ to find the embedding of SU($n$) $\supset \text{Spin}(10)?$ Is that $n \geq 7$?
How do we find such an embedding by an explicit construction?
Edit: It appears that my answer below is wrong since these half-spinor representations are not faithful. I will sort this out when I have more time to think about it.
First of all, it suffices to study complex representations $\rho: Spin(n, {\mathbb C})\to SL(N, {\mathbb C})$ of complex Spin groups $Spin(n, {\mathbb C})$: The restriction of such a representation to the compact Spin subgroup $Spin(n)$ is automatically unitarizable, i.e. the image is contained in a conjugate of $SU(N)$.
A representation $\rho$ is called spinoral (or simply spin) if it does not descend to the orthogonal group $SO(n, {\mathbb C})$ (equivalently, $\rho$ is injective). Confusingly, there are also half-spin (or semi-spin) representations: They are also spinoral.
For each simple complex Lie group $G$ irreducible (finite-dimensional) complex linear representations $G\to GL(V)$ are parameterized by weights $\lambda$; one writes $V=V(\lambda)$ in this situation (the notation $\rho$ is suppressed). Each weight $\lambda$ is the sum of fundamental weights $\omega_1,...,\omega_\ell$, where $\ell$ is the rank of the group $G$. For $G=Spin(n, {\mathbb C})$ ($n\ge 5$), $n=2\ell$ or $n=2\ell+1$ depending on the parity of $\ell$. Pierre Deligne in his Notes on spinors observes that among fundamental representations (i.e. representations whose weights are fundamental) spin-representations correspond to the nodes on the right of the Dynkin diagram: There is either one such node (labeled $\omega_\ell$, if $n$ is odd) or two ($\omega_{\ell-1}, \omega_\ell$, if $n$ is even). In the case of two nodes the corresponding representations are half-spin and they have the same dimension. Thus, it suffices to consider only $\omega_\ell$ regardless of the parity of $n$. One can verify that the corresponding representations $V(\omega_\ell)$ are the spin representations of the lowest dimension. These representations are also minuscule (the reason they have the lowest dimension among all spin-representations) and their dimensions are computed as: $$ N=dim(V(\omega_\ell))=2^{\ell}, n=2\ell+1, $$ and, in the half-spin case:
$$ N=dim(V(\omega_{\ell}))= 2^{\ell-1}, n=2\ell. $$
These are the lowest dimensions in which $Spin(n)$ embeds in $SU(N)$. For $n=10=5\times 2$ (the case you asked about) we get $N=2^{5-1}=16$, higher than you expected.
As for explicit constructions of such representations, you can find them in this Wikipedia article or in Deligne's notes.