Support of an infinitely divisible measure.   Hello, 
if $G$ is a compact Lie group. Let $\mu$ be an infinitely divisible measure on $G$, such that $e$, the neutral element of $G$, is in the support of $\mu$. Is that true that the support of $\mu $ is a group ? Why ? 
Thanks a lot. 
 A: Have you tried to look up
Heyer, Herbert, Probability measures on locally compact groups. Ergebnisse der Mathematik und ihrer Grenzgebiete. 94. Berlin-Heidelberg-New York: Springer-Verlag. (1977)?
I don't have access to this book right now, but I expect that you can find the answer there.
A: Actually Martin the support in your exemple would be the closure of $\{\exp(ix), x\in \mathbb{N}\}$, which is equal to $U(1)$, which is a group. 
In a finite group the proof would be just that if $x$ is in the support of $\mu$, looking at $\mu$ like more or less the measure of $X_1$ with $X$ a Levy process going from $e$, then it is obvious that we can "speed" up the jumps and so $x$ is in $X_{\frac{1}{2}}$, and so as the support of $X_1$ is equal to the closure of the product of the support of $X_{\frac{1}{2}}$ with himself then $\text{Support}(X_1)$ is stable by multiplication. Thus it is a group as a subset of a finite group which is stable by multiplication is a group. But in a Lie group can we "speed up" the jumps ? If $X$ is a Levy process beginning from $e$ having at any time $e$ in his support, is it true that : $Support(X_{1}) \subset Support(X_{\frac{1}{2}})$ ? 
A: Hello,
Let consider a pure jump process on the unitary group on the complex plane :
the jumps $exp(i)$ arrive randomly according to a Poisson process ;
the support is composed of $exp(ix)$ with $x\in \mathbb{N}$.
Is it a counter-example or I forget something ?
