infinite dihedral group

• Feb 14th 2010, 07:31 AM
dreamon
infinite dihedral group
"The infinite dihedral group
Dis 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?
• Feb 14th 2010, 10:12 PM
NonCommAlg
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 $tst=s$ and thus $st=t^{-1}s,$ which implies that $st^i=t^{-i}s,$ for all $i \in \mathbb{Z}.$ therefore $D_{\infty}=\{t^is^j: \ i \in \mathbb{Z}, \ j \in \{0,1\} \}.$ for the second part of your question, to find the orbit of $a \in \mathbb{R},$

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

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

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

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