Fundamental group of $U(n) / \mathbb{Z}_m$ I would like to know the fundamental group of $U(n) / \mathbb{Z}_m$, where the $\mathbb{Z}_m$ in the quotient is the diagonal matrices with entries equal to an $m$th root of unity. 
The strategy I've adopted so far is to write down the long exact sequence in homotopy groups coming from $1 \to \mathbb Z_m \to U(n) \to U(n)/\mathbb Z_m \to 1$. Since $\pi_1(U(n)) = \mathbb Z$, the relevant part of the long exact sequence is
$$ 1 \to \mathbb{Z} \xrightarrow{\pi} \pi_1(U(n)/\mathbb Z_m) \to \mathbb Z_m \to 1, $$
where the first entry is $\pi_1(\mathbb Z_m) = 1$ and the last is $\pi_0(U(n))=1$. 
As this point I've gotten stuck, but I can at least say the following. First, I believe the map $\pi$ is multiplication by $m / \gcd(m,n)$, which comes from looking at how the generator of $\pi_1(U(n))$, which is determined by the determinant map, behaves when the $\mathbb Z_m$ quotient is taken. Additionally, when $m=n$ I think that $\pi_1(U(n)/\mathbb Z_n)) = \mathbb Z \oplus \mathbb Z_n$, while if $\gcd(m,n)=1$ then $\pi_1(U(n)/\mathbb Z_m)) = \mathbb Z$. It also seems reasonable to me that the torsion part of $\pi_1(U(n)/\mathbb Z_m)$ should be $\mathbb Z_{\gcd(m,n)}$ in general, with the torsion coming from when the $\mathbb Z_m$ quotient hits the $SU(n)$ factor in $U(n)$. 
 A: Your guesses were correct.  We will show:

The quotient $U(n)/\mathbb{Z}_m$ has fundamental group isomorphic to $\mathbb{Z}\oplus \mathbb{Z}_{\gcd(n,m)}.$

First, $U(n)$ is diffeomorphic (but not Lie isomorphic) to $SU(n)\times S^1$ via the map $f:U(n)\rightarrow SU(n)\times S^1$ given by $f(A) = ( \operatorname{diag}(\overline{\det(A)}, 1,..., 1) A, \det A).$  The inverse is $g:SU(n)\times S^1\rightarrow U(n)$ given by $g(A,z) = \operatorname{diag}(z,1,...,1) A$.
We can use $f$ and g to transfer the $\mathbb{Z}_m$ action to $SU(n)\times S^1$ in such a way as to make $f$ and $g$ equivariant diffeos.  For $\lambda \in \mathbb{Z}_m$, we define \begin{align*}\lambda \cdot (A,z) &:= f(\lambda\cdot g(A,z))\\ &= f(\lambda \operatorname{diag}(z,1,...,1) A)\\ &= (\operatorname{diag}(\overline{z\lambda^n}, 1,....,1) \lambda \operatorname{diag}(z,1,...,1) A, z\lambda^n)\\&= (\operatorname{diag}( \lambda^{-n+1},\lambda,....,\lambda)  A, z\lambda^n).\end{align*}
The advantage of this description is that it makes it easier to see the corresponding action on the universal cover $SU(n)\times \mathbb{R}$.
Proposition:  Let $f,g:SU(n)\times \mathbb{R}\rightarrow SU(n)\times \mathbb{R}$ with $f(A,t) = (A, t + 2\pi)$ and $g(A,t) = \left(\operatorname{diag}(\lambda^{-n+1}, \lambda,...,\lambda) A, t + \frac{2\pi n}{m}\right)$, where $\lambda = e^{2\pi i /m}$.  Suppose $G$ is the group generated by $f$ and $g$.  Then $G$ is the deck group of the covering $SU(n)\times \mathbb{R}\rightarrow (SU(n)\times S^1)/ \mathbb{Z}_m$.
Proof:  Because $f$ and $g$ obviously commute, we can first quotient by the $f$ action and then the $g$ action.  Now, the covering map $\pi:SU(n)\times \mathbb{R}\rightarrow (SU(n)\times S^1)/\mathbb{Z}_m$ factors as the composition $$SU(n)\times \mathbb{R}\rightarrow SU(n)\times S^1\rightarrow (SU(n)\times S^1)/\mathbb{Z}_m.$$
The first map of this composition is the quotient $$(SU(n)\times \mathbb{R}\rightarrow (SU(n)\times \mathbb{R})/\langle f\rangle\cong SU(n)\times S^1$$ and the second is the map $$SU(n)\times S^1\rightarrow (SU(n)\times S^1)/\langle g\rangle \cong (SU(n)\times S^1)/\langle \mathbb{Z}_m\rangle.$$ $\square$
It follows from this proposition that $G\cong \pi_1( (SU(n)\times S^1)/\mathbb{Z}_m)$.  So let's figure out what $G$ is, exactly.
Proposition:  The group $G$ is isomorphic to $\mathbb{Z}\oplus \mathbb{Z}_{\gcd(a,b)}.$
Proof:  Since $G$ is generated by the commuting maps and $f$ and $g$, there is a surjection $\phi:\mathbb{Z}^2\rightarrow G$ with $\phi(1,0) = f$ and $\phi(0,1) = g$.  We claim $\ker \phi = \langle (-n,m) \rangle$.
To see this, first note that $g^{m}$ fixes all points in the $SU(n)$ factor since $\lambda \in \mathbb{Z}_m$.  Further, $g^{m}$ translates the $\mathbb{R}$ factor by $m \cdot \frac{2\pi n}{m} = 2\pi n$.  Since $f^{-n}$ translates the opposite direction by $2\pi n$, $f^{-n}g^{m} = 1$.  This shows that $\langle (-n,m)\rangle \subseteq \ker \phi$.
On the other hand, if $f^\alpha g^\beta = 1$.  Then $\beta = k m$ for some integer $k$ since otherwise $f^\alpha g^\beta$ moves points in $SU(n)$.  Then $g^\beta$ translates the $\mathbb{R}$ factor by $2\pi nk$ steps, so we must have $\alpha = - nk$.  Thus, $\ker \phi \subseteq \langle (-n,m)\rangle$.
Thus, by the first isomorphism theorem, $G\cong \mathbb{Z}^2/\langle (-n,m)\rangle$.  Finally, from, e.g., this MSE answer of Martin Brandeburg, this is isomorphic to $\mathbb{Z}\oplus \mathbb{Z}_{\gcd(n,m)}.$ $\square$
