When we have a Free Product G between two groups G1 G2 we can construct a Tree, such that G1*G2 acts on the edges without inversion, transitive and free. The tree has vertices in G/G1 and G/G2. The positiv oriented edges are in G. We set the beginning of an edge to be the edge multiplied with G1 and the end to be the edge multiplied with G2.
Now I want to show that there are no nontrivial reduced paths. (I proof the theorem like Oleg Bogopolsky. At the end I added a detail but my teacher said this was unnecessary and the proof was already complete earlier.)Suppose there is one closed nontrivial reduced path. Then without restriction the path starts at a coset of G1. We can also assume that the path starts at G1 (we can multiply every edge with an Element and the path stays closed and reduced). Since edges jump between G1 and G2 there have to be even edges in the path.
Now we know that for xi in G1 yi in G2 the end of e1 is x1G2 e2 is x1y1G1 en is x1y1...x(n/2)y(n/2)G1 = G1
So we know that x1...y(n/2) is in G1 so all yi must be 1.(my teacher says i can stop here.)
I still continued my proof: So y1 is 1. Now i want to show that x1 is the only edge from G1 to x1G2. Then the edge e2 from x1G2 to x1y1G1 = G1 must be the inverse. Let x1G1 = gG1 and x1G2 = gG2 then since the intersection of G1 and G2 is the neutral element, we conclude x1 = g.
Why is this last step unnessecary?