The following situation is given: Let $n>1$ be a natural number and consider the group $\mathcal{U}_n(\mathbb{C})=\{S\in M_n(\mathbb{C})| \text{S unitary}\} $ and $\mathrm{Tr}:M_n(\mathbb{C})\to \mathbb{C},\; (a_{ij})\mapsto \sum\limits_{j=1}^na_{jj}$ the trace. Consider $$S\mathcal{U}_n^{\mathrm{Tr}}(\mathbb{C})=\{e^{ih_1}\cdots e^{ih_k}| \; k\in\mathbb{N},\;h_1,\cdots h_k\in M_n(\mathbb{C})_{sa},\; \frac{1}{n}\mathrm{Tr}(\sum\limits_{j=1}^kh_{j})=0\}.$$ $M_n(\mathbb{C})_{sa}$ are the matrices in $M_n(\mathbb{C})$ which are self-adjoint. $S\mathcal{U}_n^{\mathrm{Tr}}(\mathbb{C})$ defines a normal subgroup of $\mathcal{U}_n(\mathbb{C})$
What is $$\mathcal{U}_n(\mathbb{C})/S\mathcal{U}_n^{\mathrm{Tr}}(\mathbb{C}),$$can we characterize/describe this factor group or identify it with a (well) known group? If so, how? Thank you.
Your group $S\mathcal{U}_n^{\mathrm{Tr}}(\mathbb{C})$ is just the group $SU(n)$ of unitary matrices with determinant $1$. Indeed, if $u=e^{ih_1}\cdots e^{ih_k}\in S\mathcal{U}_n^{\mathrm{Tr}}(\mathbb{C})$, then $$\det(u)=\prod \det(e^{ih_j})=\prod e^{i\mathrm{Tr}(h_j)}=e^{i\sum \mathrm{Tr}(h_j)}=e^{i\mathrm{Tr}\left(\sum h_j\right)}=1.$$ Conversely, if $u\in SU(n)$, then $u$ is diagonalizable. Choosing logarithms for each of the eigenvalues of $u$, we get a self-adjoint matrix $h$ such that $e^{ih}=u$ and the sum of the eigenvalues of $h$ is a multiple of $2\pi$. Subtracting that multiple from one of the eigenvalues, we may assume the sum is actually $0$, so $\mathrm{Tr}(h)=0$ and $u=e^{ih}\in S\mathcal{U}_n^{\mathrm{Tr}}(\mathbb{C})$.
In particular, $SU(n)$ is the kernel of the homomorphism $\det:U(n)\to \mathbb{C}^\times$. Thus the quotient $U(n)/SU(n)$ is isomorphic to the image of $\det$, which is just the group $S^1\subset \mathbb{C}^\times$ of complex numbers of norm $1$.