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: $A*_{A,Id_{A}}=\langle a_{i},t|R(a_{i})=1,t^{-1}a_{s}t=Id_{A}(a_{s})=a_{s}\rangle$.

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

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