Quote:

Originally Posted by

**dreamon**

"The infinite dihedral group D∞ is generated (as a subgroup of the group SR of bijections : R ! R), by the translation t(x) = x + 1 and the reflection s(x) = −x of the real line. Work out its elements, and find the orbit and the stabilizer of each of the points 1, 1/2, 1/3".

I have to deal with that problem but i have no idea that how i am going to use that points. I once read that ts(x)= -x+1 has reflection in the point 1/2 but i dont know how.

Anyone to explain?

first see that $\displaystyle tst=s$ and thus $\displaystyle st=t^{-1}s,$ which implies that $\displaystyle st^i=t^{-i}s,$ for all $\displaystyle i \in \mathbb{Z}.$ therefore $\displaystyle D_{\infty}=\{t^is^j: \ i \in \mathbb{Z}, \ j \in \{0,1\} \}.$ for the second part of your question, to find the orbit of $\displaystyle a \in \mathbb{R},$

we need to find $\displaystyle t^is^j(a),$ for any $\displaystyle i \in \mathbb{Z}, \ j \in \{0,1\}.$ we have: $\displaystyle s^0(a)=a, \ s(a)=-a.$ so we can write $\displaystyle s^j(a)=(-1)^ja.$ thus $\displaystyle t^is^j(a)=t^i((-1)^ja)=(-1)^ja + i.$ therefore the orbit of $\displaystyle a$ is the set

$\displaystyle \{(-1)^ja + i : \ i \in \mathbb{Z}, \ j \in \{0,1 \} \}.$ in order to find he stabilizer of $\displaystyle a$ we need to find all $\displaystyle i \in \mathbb{Z}, \ j \in \{0,1 \}$ such that $\displaystyle (-1)^ja+i=a.$ that will depend on $\displaystyle a.$ for example, if $\displaystyle a=1,$ then we need to

solve $\displaystyle (-1)^j + i = 1.$ if $\displaystyle j=0,$ then we have $\displaystyle i = 0$ and if $\displaystyle j=1,$ then $\displaystyle i=2.$ so the stabilizer of $\displaystyle a=1$ is $\displaystyle \{1, t^2s \}.$

for $\displaystyle a=1/2$ we need to solve $\displaystyle (-1)^j/2 + i= 1/2,$ which can be written as $\displaystyle (-1)^j + 2i = 1.$ now if $\displaystyle j=0,$ then $\displaystyle i=0$ and if $\displaystyle j=1,$ we'll get $\displaystyle i=1.$ so the stabilizer of $\displaystyle a=1/2$ is $\displaystyle \{1,ts \}.$