Group of distance preserving transformations of the plane is isomorphic to $\mathbb{Z} \ltimes \mathbb{Z}$ "Let $G$ be the group of distance preserving transformations of $\mathbb{R^2}$ which is generated by $(x,y)\mapsto(x+1,y)$and $(x,y)\mapsto(-x,y+1)$.
Prove that $G$ is isomorphic to the semidirect product $\mathbb{Z} \ltimes \mathbb{Z}$ where f sends $1$ to the non-trivial automorpism of $\mathbb{Z}$."
From "Groups and Symmetry" by M.A. Armstrong, 23.12.
I'm thoroughly confused; I think I don't really get how automorphisms work.
The nontrivial automorphism is obviously $f(1)=-1$. So the multiplication would presumably be:
$(x,y)(x',y')=(x.f(y)(x'),yy')$ 
However here I'm not sure what this means; $f$ takes $y$ to $-y$ so presumably we'd end up with:
$(x,y)(x',y')=(-xyx',yy')$ 
I've written some attempts at proving injectivity and surjectivity but I've a strong feeling  I'm misunderstanding automorphisms completely, and it is something different all together. 
Any help would be much appreciated.
 A: There are several points of confusion here.
First, as addressed in the comments, the group operation on $\mathbb{Z}$ is addition, not multiplication.
Second, it is not true that $f$ takes $y$ to $-y$.  Remember that $f$ is a map from $\mathbb{Z}$ to Aut($\mathbb{Z}$) (automorphisms of $\mathbb{Z}$), not a map from $\mathbb{Z}$ to $\mathbb{Z}$.  When you write $f(1) = -1$, that is correct only if you interpret "-1" as the automorphism "multiplication by -1".  The output of $f$ has to be an automorphism of $\mathbb{Z}$, not an integer: Given $y \in \mathbb{Z}$, to define $f(y)$, you need to define $f(y)(x)$ for every $x \in \mathbb{Z}$.  So when you write $f(1) = -1$, that really means $f(1)(x) = -x$, i.e., $f(1)$ multiplies any $x \in \mathbb{Z}$ by -1.
So the first step is to work out what $f(y)$ is for any $y \in \mathbb{Z}$.  Remember, you know that $f(1)(x) = -x$, and $f$ must be a homomorphism from $\mathbb{Z}$ to Aut($\mathbb{Z}$).  Both $\mathbb{Z}$ and Aut($\mathbb{Z}$) are groups, but while the group operation on $\mathbb{Z}$ is addition, the group operation on Aut($\mathbb{Z}$) is function composition.
To make the point in another way: It is not correct to say that $$f(2) = f(1+1) = f(1) + f(1) = -1 + -1 = -2 \rm{\mbox{  (INCORRECT)}}$$ because the "+" in the "$f(1)+f(1)$" term should be the group operation on Aut($\mathbb{Z}$), i.e., function composition. You can also convince yourself that there is no such automorphism as -2 (multiplication by -2 is not an automorphism of $\mathbb{Z}$ so it couldn't be an output of $f$.)
Once you work out what $f(y)$ is, you should be able to write down the correct formula for the group operation in the semidirect product.
A: Letting $\,\phi:\Bbb Z\to\operatorname{Aut}(\Bbb Z)\,\,,\,\,\phi(1):=\psi\,$ , where $\,\psi(m):=-m\,\,,\,\,\forall\,m\in\Bbb Z\,$ , we get:
$$\phi(k):=\psi^k=\begin{cases}\psi&,\,\;\;k\,\,\text{is odd}\\Id_{\Bbb Z}&,\;\;\,k\,\,\text{ is even}\end{cases}$$
so that we  have
$$\Bbb Z\rtimes\Bbb Z:=\left\{(a,b)\in\Bbb Z\times\Bbb Z\;\;;\;\;(a,b)*(a',b'):=\left(a+a'^{\phi_b},b+b'\right)=\left(a+(-1)^ba',b+b'\right)\right\}$$
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 $\,S^{-1}TS=T^{-1}\,$ , so perhaps you can take it from here...
