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UTM Armstrong groups and symmetry Exercise 23.12
Let $G$ be the group of distance preserving transformations of $\mathbb{R}^2$ which is generated by $(x, y) \rightarrow(x+1, y)$ and $(x, y) \rightarrow(-x, y+1)$. Prove that $G$ is isomorphic to the semidirect product $\mathbb{Z} \times_{\varphi} \mathbb{Z}$ where $\varphi$ sends 1 to the non-trivial automorphism of $\mathbb{Z}$.
non-trivial automorphism of $\mathbb{Z}$就是$\psi(n)=-n$ 所以$\varphi(1)=\psi$
$\psi$的$n$次复合为$\psi^m(n)=(-1)^mn$, 因为$\varphi$是$\mathbb Z$到$\operatorname{Aut}\mathbb Z$的同态, 所以$\varphi(m)(n)=(-1)^mn$
我们有
$$\Bbb Z\times_\varphi\Bbb Z:=\left\{(a,b)\in\Bbb Z\times\Bbb Z\;\;;\;\;(a,b)*(a',b'):=\left(a+\varphi(b)(a'),b+b'\right)=\left(a+(-1)^ba',b+b'\right)\right\}$$
Generators of this group are
$$\left\{\alpha:=(1,0)\,,\,\beta:=(0,1)\right\}$$
with relations $\beta^{-1}\alpha\beta=\alpha^{-1}$
Take now the distance-preserving (rigid) transformations of the plane
$$T(x,y):=(x+1,y)\,,\,S(x,y):=(-x,y+1)\Longrightarrow $$
$$T^{-1}(x,y):=(x-1,y)\,,\,S^{-1}(x,y):=(-x,y-1)$$
Note that $\,S^{-1}TS=T^{-1}$,
$a\mapsto T,b\mapsto S$ is a group isomorphism:
$$f(a,b)f(a',b')=((x+a)(-1)^b,y+b)((x+a)(-1)^{b'},y+b')=f(aa',bb')$$
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