User:Juan Marquez/HNN-extensions

Z_2*_{Z_2}Z_2
am I sure that $$\mathbb{Z}_2*_{\mathbb{Z}_2}=\mathbb{Z}_2*_{\mathbb{Z}_2}\mathbb{Z}_2$$ is equal to $$\mathbb{Z}$$? Not yet...

but, it is solved so...
We have for the HNN-extension: $$A*_{A,Id_A}=\langle a_i,t|R(a_i)=1,t^{-1}a_st=Id_A(a_s)=a_s\rangle$$.

Which in the case $$A=\mathbb{Z}_2$$ will give us
 * $$\mathbb{Z}_2*_{\mathbb{Z}_2}=

\langle a,t|a^2=1,t^{-1}at=a\rangle= \mathbb{Z}_2\oplus\mathbb{Z}$$

For the trivial homomorphism $$0:\mathbb{Z}_2\to{\mathbb{Z}_2} ;\ x\mapsto 1$$ we have
 * $$\mathbb{Z}_2*_{\mathbb{Z}_2,0}=

\langle a,t|a^2=1,t^{-1}at=1\rangle=\mathbb{Z}$$

Which shows indeed that $$\mathbb{Z}$$ is factorizable within finite groups

oops!
In fact that we've just seen is that $$\mathbb{Z}= \mathbb{Z}_2*_{\mathbb{Z}_2}\mathbb{Z}_2$$ because we have used the only two group-morphism $$\mathbb{Z}_2\to\mathbb{Z}_2$$ in the definition of amalgamated free product

symbolic
$$\scriptstyle G=\langle H, t | t^{-1}Kt=L\rangle$$