# Volume factor in Faddeev-Popov quantisation

In Faddeev-Popov quantisation, why does the integral over gauge parameter cancel the volume factor of the gauge group that's in the denominator? In fact, I don't understand where the volume factor comes from at the first place.

$Z = \int d [A] e^{iS}$
and since it has a gauge symmetry, there are multiple values of $A$ that generating the same Action $S$, since $S(A_g)=S(A_{g'})$ for the two different choices of gauge $g,g'$.
Now you have a gauge-fixing condition like $\partial^\mu A_\mu = 0$. The integral which contains enough physical Information is than given by $Z = \int d [A] \delta(A_\mu|_{gauge-fixed}) e^{iS}$ (1), but unfortunately you have the integral $\int d [A] \delta(\partial^\mu A_\mu)e^{iS}$ (2). Recall the identity for delta-distributions $\delta(f(x)) = \sum_{i \in roots(f)} \frac{\delta(x-x_i)}{|f'(x_i)|}$ and generalize it to the "infinite-dimensional" function $A$. You have a Jacobian determinant in the denominator. But when you provide an additional factor in the Integrand of (2) which is simply the Jacobian functional determinant, it would cancel out and you would arrive on the integral (1).