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The question entails that I should choose two finite groups, then construct a 'biregular' tree, and show that the action of the free product of the two finite groups on the biregular tree will have a fundamental domain that consists of a single edge and two vertices. What I have so far is the two finite groups. The first group is $A = C_2$, and the second group is $B = D_4$. I know the group presentations of these groups. I understand that the free product of A and B is a group of symmetries of the biregular tree. I am lost on how to construct the biregular tree. If anyone can offer some suggestions or help it would be greatly appreciated. Thanks in advance.

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You graph should consist of (infinitely many) vertices with $4$ incident edges and (infinitely many) vertices with $2$ incident edges, joined in such a way that to get from any vertex with valency $4$ to any other with valency $4$ you have to travel through a vertex with valency $2$. That is, your graph consists of vertices with valency $4$ connected only to vertices with valency $2$, and vertices with valency $2$ connected only to vertices with valency $4$. $C_2$ acts on the vertices with valency $2$, while $D_4$ acts on the vertices with valency $4$. –  user1729 Jan 15 '12 at 20:17
    
Thank you very much! –  Aggie Jan 16 '12 at 0:44

1 Answer 1

To make my comment an answer:

You graph should consist of (infinitely many) vertices with 4 incident edges and (infinitely many) vertices with 2 incident edges, joined in such a way that to get from any vertex with valency 4 to any other with valency 4 you have to travel through a vertex with valency 2. That is, your graph consists of vertices with valency 4 connected only to vertices with valency 2, and vertices with valency 2 connected only to vertices with valency 4. The given copy of $C_2$ acts on a fixed vertex with valency 2 in the natural way, while the given copy of $D_4$ acts on an adjacent vertex with valency 4 in the natural way. These actions extend to the whole tree, again in the natural way.

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