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**Banach** Thanks for your answer but i unfortunately i am not familiar with these terms. However, i think i know how to work it out now:

Wlog we may assume that $\displaystyle F=F(a,b)$. Let $\displaystyle \Gamma_1$ and $\displaystyle \Gamma_2$ the associated graphs of G and H. Define the graph $\displaystyle \Gamma$ by $\displaystyle V(\Gamma):=V(\Gamma_1) \times V(\Gamma_2)$ and there is an edge between (a,x) and (b,y) iff there are edges from a to b in $\displaystyle \Gamma_1$ and x to y in $\displaystyle \Gamma_2$ having the same label. Since G and H are f.g. the graphs $\displaystyle \Gamma_1$ and $\displaystyle \Gamma_2$ are finite graphs, thus $\displaystyle \Gamma$ is also a finite graph whose fundamental group is $\displaystyle G \cap H$. Hence the intersection is f.g., qed.

What do you think of that proof?

Greetings

Banach