Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

"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.

share|improve this question
2  
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

2 Answers 2

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.

share|improve this answer
    
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

$$\left\{\alpha:=(1,0)\,,\,\beta:=(0,1)\right\}$$

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...

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.