By Clara Löh

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**Example text**

10 (Free actions on Cayley graphs). Let G be a group and let S be a generating set of G. Then the left translation action on the Cayley graph Cay(G, S) is free if and only if S does not contain any elements of order 2. Recall that the order of a group element g of a group G is the infinimum of all n ∈ N>0 with g n = e; here, we use the convention inf ∅ := ∞. Proof. The action on the vertices is nothing but the left translation action by G on itself, which is free. It therefore suffices to study under which conditions the action of G on the edges is free: If the action of G on the edges of the Cayley graph Cay(G, S) is not free, then S contains an element of order 2: Let g ∈ G, and let {v, v } be an edge of Cay(G, S) with {v, v } = g · {v, v } = {g · v, g · v }; by definition, we can write v = v · s with s ∈ S ∪ S −1 \ {e}.

But the family of BaumslagSolitar groups also contains many intriguing examples of groups. , there exists a surjective group homomorphism BS(2, 3) −→ BS(2, 3) that is not an isomorphism [9], namely the homomorphism given by BS(2, 3) −→ BS(2, 3) a −→ a2 b −→ b. However, proving that this homomorphism is not injective requires more advanced techniques. 2. 23 (complicated trivial group). The group G := x, y | xyx−1 = y 2 , yxy −1 = x2 is trivial because: Let x ∈ G and y ∈ G denote the images of x and y respectively under the canonical projection F ({x, y}) −→ F ({x, y})/ {xyx−1 y −2 , yxy −1 x−2 } F (S) = G.

1. The above composition is well-defined because if two reduced words are composed, then the composed word is reduced by construction. ) as neutral element, and it is not difficult to show that every reduced word admits an inverse with respect to this composition (take the inverse sequence and flip the bar status of every element). 3. 8): Let x, y, z ∈ Fred (S); we want to show that (x · y) · z = x · (y · z). By definition, when composing two reduced words, we have to remove the maximal reduction area where the two words meet.

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