Why $\{e\} \ast_{\mathbb{Z}} \mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$?

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Am I correct? Following Serre's "Trees", I can conclude that $\{e\}\ast_{\mathbb{Z}} \mathbb{Z} = \varinjlim (\{e\},\mathbb{Z},\mathbb{Z})$ with respect to homomorphisms:

a) $\mathbb{Z} \mapsto e$,

ii) $\mathbb{Z} \to \mathbb{Z}$ given by $a \mapsto -a$,

в) $e \mapsto 0 \in \mathbb{Z}$.

So we have $\varinjlim (\{e\},\mathbb{Z},\mathbb{Z}) = (\{e\} \oplus \mathbb{Z} \oplus \mathbb{Z})/\langle(e,a,0)\equiv(e,0,0), (e,0,b) \equiv (e,0,-b) \rangle_{a,b \in \mathbb{Z}}$ which is obviously $\mathbb{Z}/2$.