The disintegration of the convolution of two probability measures 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?
 A: $\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\si}{\sigma}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$
The answer is no, in general it is not possible to express the measure $(\mu * \nu)_h$ in terms of the families $(\mu_h)_{h \in H}$ and $(\nu_h)_{h \in H}$ only. 
E.g., suppose that $G$ is the additive group $\R^2$, $N=\R\times\{0\}$, measures $\mu,\hat\mu,\nu$ are probability measures, 
$\bullet\quad$ $\mu$ attaches masses $\frac28,\frac18,\frac58$ to points $(0,0),(0,1),(1,0)$, respectively;
$\bullet\quad$ $\hat\mu$ attaches masses $\frac24,\frac14,\frac14$ to points $(0,0),(0,1),(1,0)$, respectively; 
$\bullet\quad$ $\nu$ attaches masses $\frac14,\frac14,\frac24$ to points $(0,0),(0,1),(1,0)$, respectively.  
Then $\mu_h=\hat\mu_h$ for $p_* \mu$-almost all $h \in H$ and, equivalently, for $p_* \hat\mu$-almost all $h \in H$. 
However, 
$$(\mu * \nu)_1(\{0\})=\frac9{16}\ne\frac58=(\hat\mu * \nu)_1(\{0\}).$$ 

Added in response to the OP's modification of the original question by adding the semigroup requirement:
Let $G$ and $H$ be as above. Let $(A_t)$, $(B_t)$, $(C_t)$, $(D_t)$ be independent standard Brownian motions.
For $t\ge0$, let 
\begin{equation}
 X_t:=\si_1 A_t,\quad Y_t:=\si_2(\rho A_t+\sqrt{1-\rho^2}B_t),
\end{equation}
where 
\begin{equation}
 \si_2:=\frac1{\sqrt{1-\rho^2}},\quad \si_1:=\rho\si_2=\frac\rho{\sqrt{1-\rho^2}}, 
\end{equation}
and $\rho\in(0,1)$. 
For each $t\ge0$, let $\mu_t$ and $\nu_t$ be the probability distributions of $(X_t,Y_t)$ and $(C_t,D_t)$, respectively, so that $\mu_t$ and $\nu_t$ are the bivariate normal distributions $N(0,0,\si_1^2 t,\si_2^2 t,\rho)$ and $N(0,0,t,t,0)$, and hence 
\begin{align*}
 \mu_t * \nu_t&=N\Big(0,0,(\si_1^2+1)t,(\si_2^2+1)t,\frac{\rho\si_1\si_2}{\sqrt{\si_1^2+1}\sqrt{\si_2^2+1}}\Big) \\ 
 &=N\Big(0,0,\frac t{1-\rho^2},\frac{(2-\rho^2)t}{1-\rho^2},
 \frac{\rho^2}{\sqrt{2-\rho^2}}\Big). 
\end{align*}
Moreover, obviously $(\mu_t)$ and $(\nu_t)$ are semigroups. 
Next, for each $h\in\R$ ($\R$ being identified with $H=\R\times\{0\}$), the measure $(\mu_t)_h$ is the conditional distribution of $Y_t$ given $X_t=h$, so that (cf. e.g. page 4 of bivariate normal distribution ) 
\begin{equation}
 (\mu_t)_h=N\Big(\rho\frac{\si_2}{\si_1}\,h,(1-\rho^2)\si_2^2 t\Big)
 =N(h,t),
\end{equation}
which does not depend on $\rho$. 
However, 
\begin{equation}
 (\mu_t *\nu_t)_h=N(\rho\sqrt{2-\rho^2}\, h,(2+\rho^2)t)
\end{equation}
obviously does depend on $\rho$.  
Thus, $(\mu_t *\nu_t)_h$ is not determined by $(\mu_t)_h$ and $(\nu_t)_h$, for any real $h$ and any $t>0$. 
A: The easiest is to look just at the situation when the quotient measures on $G/H$ are purely atomic, so that
$$
\mu = \sum_h \alpha_h \mu_h
$$
and
$$
\nu = \sum_h \beta_h \nu_h \;,
$$
where $\alpha$ and $\beta$ are two discrete probability distributions on the quotient group $H$, and $\mu_h,\nu_h$ are probability measures on the fibers $p^{-1}(h)$. Then 
$$
\mu*\nu = \sum_{h,g} \alpha_h \beta_g \, \mu_h*\nu_g \;,
$$
whence 
$$
(\mu*\nu)_x = \frac{\sum_{h,g:hg=x} \alpha_h \beta_g \, \mu_h*\nu_g}{\sum_{h,g:hg=x} \alpha_h \beta_g} \;.
$$
For the simplest example then take
$$
\alpha = \alpha_1 \delta_{h_1} + \alpha_1 \delta_{h_2} 
$$
and
$$
\beta = \beta_1 \delta_{h_1^{-1}} + \beta_2 \delta_{h_2^{-1}} \;.
$$
Then
$$
(\mu*\nu)_e = \frac{\alpha_1\beta_1 \, \mu_{h_1}*\nu_{h_1^{-1}} + \alpha_2\beta_2 \, \mu_{h_2}*\nu_{h_2^{-1}}}{\alpha_1\beta_1 + \alpha_2\beta_2} \;,
$$
which obviously depends on the weight distributions $\alpha$ and $\beta$.
