Let $G$ be a topological group with all the topological conditions in order that some form of the disintegration theorem be applicable (for instance, take $G$ metrizable). Let $N$ be normal and closed, and let $p : G \to G/N = H$ be the natural projection.

Let $\mu$ be a regular Borel probability. Disintegrating $\mu$ with respect to $p_* \mu$, one gets a family $(\mu_h)_{h \in H}$ of regular Borel probabilities on $G$, concentrated on the fibre $p^{-1} (h)$ for $p_* \mu$-almost all $h \in H$, such that $\mu (B) = \int _H \mu_h (B) \ \mathrm d (p_* \mu) (h)$ for all Borel $B \subseteq G$.

With notations as above, is it possible to express the measure $(\mu * \nu)_h$ in terms of the families $(\mu_h)_{h \in H}$ and $(\nu_h)_{h \in H}$?

In the beginning, I was expecting this to be a trivial one-line warm-up calculation. To my frustration, this didn't really happen: either I am overlooking something, or such a simple formula does not exist.

To give a clearer view of what I am interested in, imagine that $(\mu_t)_{t \ge 0}$ is a semigroup of probabilities on $G$. Since $(p_* \mu_t)_{t \ge 0}$ is a semigroup on $H$, I found it natural to ask whether the semigroup property is transmitted in any way (not necessarily as a semigroup property again) to the fibers. Semigroup on thw whole space, semigroup on the base of the fibration - wouldn't it be natural to expect something "nice" on the fibers, too?