In my PDE course we had a short section on Brownian motion. Let $G_t(x):= (4\pi t)^{-\frac{n}{2}} e^{-\frac{(x-y)^2}{4t}}$ denote the heat kernel. Then $P(t,x,y):=G_t(x-y)$ is an example for a Brownian motion. For $f\in BUC(\mathbb{R}^n)$ we have that $(T(t)f)(x):= \int P(t,x,y) f(y)\,\mathrm{dy}$ defines a $C_0$ semigroup on $BUC(\mathbb{R}^n)$.
Then we have a theorem saying that the semigroup (defined as above) of a Brownian motion has generator $c\cdot \triangle$ for some $c > 0$. Then it is stated that this means $P(t,x,y) = G_{ct}(x-y)$.
The last statement is unclear to me. My idea was: Convolution with $G_{ct}(x)$ yields a solution operator for $u_t - c\triangle u = 0$. The solution operator gives raise to a semigroup with generator $c\triangle$ which is the same as for the Brownian motion semigroup above. If the fact that the generators of the semigroups agree would imply that the semigroups are the same, then the fundamental lemma would yield $P(t,x,y) = G_{ct}(x-y)$ but I don't think that this is true, since the generator is some kind of derivative of the semigroup and in most cases derivation is not an injective process.