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"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:


However here I'm not sure what this means; $f$ takes $y$ to $-y$ so presumably we'd end up with:


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.

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The group operation on $\mathbb{Z}$ is addition, not multiplication. That may be your source of confusion. – Ted Jan 19 '13 at 19:33
@Ted I'm aware that that is the case, but I'm not sure what that would imply for the problem. Even if I change multiplication to addition like this $(x,y)(x′,y′)=(x-y+x′,y+y′)$, I'm not sure how to proceed. (if this is correct.) – Lee Wang Jan 19 '13 at 20:01
This formula is still not correct. See my answer. – Ted Jan 19 '13 at 20:34
up vote 1 down vote accepted

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.

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Thank you for this fantastic answer! This really clears up a lot of confusion for me. I've read a lot about how one should be carefull with distinguishing the function $f$ and its argument $f(x)$, would this be one of those mistakes? – Lee Wang Jan 20 '13 at 11:39

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


with relations $\,\beta^{-1}*\alpha*\beta=\alpha^{-1}\,$

Take now the distance-preserving (rigid) transformations of the plane

$$T(x,y):=(x+1,y)\,,\,S(x,y):=(-x,y+1)\Longrightarrow $$


Note that $\,S^{-1}TS=T^{-1}\,$ , so perhaps you can take it from here...

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Unfortunately I'm not able to give out two "Best answers" otherwise I surely would have. I'll take it from there: To prove the two groups isomorphic consider the function $f:= {f(a)=T,f(b)=S $ (This is not the right notation.). Now $f$ is homomorphic: $f(a,b)f(a',b')=((x+a)(-1)^b,y+b)((x+a)(-1)^b',y+b')=f(aa',bb')$. Surjectivity and injectivity are easily checked. – Lee Wang Jan 20 '13 at 11:51
@LeeWang, don't worry about the "best answer" thing: you choose whatever answer is more helpful, clearer, etc. for you. Anyway, you can upvote as many answers as you like. :) – DonAntonio Jan 20 '13 at 12:03

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