Subscript $Q$ in $U(1)_Q$ In most quantum field theory books, you read something like

QED is a QFT from Abelian $U(1)_Q$ that describes the electromagnetic interaction ...

But what does the subscript $Q$ stands for in the $U(1)_Q$ group?
 A: It specifies that one is referring specifically to the symmetry associated with electromagnetism ($Q$ stands for electric charge, but some texts also write $U(1)_{\text{EM}}$, for example). This is meant to distinguish from other possible $U(1)$ symmetries present in the theory or in related theories that correspond to other physical aspects.
For example, in the Standard Model of Particle Physics, one has the gauge group $SU(3)_{C} \times SU(2)_{L} \times U(1)_{Y}$. Notice there is a $U(1)$ factor in there. One of the interactions described by this theory is electromagnetism. However, the generator of $U(1)_Q$ (and hence the boson associated with it, which is the photon) is not the generator of $U(1)_Y$. In fact, due to a process of spontaneous symmetry breaking one ends up with a $U(1)$ symmetry whose generator is a linear combination of one generator of $SU(2)_L$ and one generator of $U(1)_Y$.
In short, sometimes there are other interesting $U(1)$ transformations in your theory, with different interpretations. To be clear about which transformation we mean, it is common to add a sub-index to the gauge group. This is the same reason $SU(3)_{C} \times SU(2)_{L} \times U(1)_{Y}$ has all these indices ($C$ means color, since it is associated with QCD; $L$ means left, since the symmetry only transforms left-handed particles; $Y$ means (weak) hypercharge, which is another sort of $U(1)$ transformation relevant in the Standard Model).
For an extra example, the neutrinos have weak hypercharge -1 (this relates to $U(1)_Y$), but they have zero electric charge (this relates to $U(1)_Q$). This is not a problem, since these are two different symmetries, both associated with a $U(1)$ group.
A: In another way of describing $Q$ as a conserved charge under $U(1)$ symmetry which could be explained by Noether's theorem.
