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Let $G$ be the set of bijections $\mathbb{R} \to \mathbb{R}$ which preserve the distance between pairs of points, and send integers to integers. Then $G$ is a group under composition of functions. The following two elements are obviously in $G$: the function $t$ (translation) where $t(x)=x+1$ for each $x \in \mathbb{R}$ and the function $r$ (reflection) where $r(x)=-x$ for each $x \in \mathbb{R}$. The subgroup of $G$ generated by $r$ and $t$ is called the infinite dihedral group and denoted by $D_{\infty}$. Note that this information describes an action of $D_{\infty}$ on $\Re$.

  1. Show that every element of $D_{\infty}$ can be written uniquely in the form $r^it^j$ for suitably restricted values of $i$ and $j$. Explain how to multiply two such elements.

  2. Describe geometrically the possible sorts of actions of elements of $D_{\infty}$ on the real line $\mathbb{R}$.

I have shown that every element is of the form $r^it^j$ but I am unsure how two elements would be multiplied together and how the actions should be described geometrically.

Can anybody help?

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I assume, you have also noted that $i\in\{0,1\}$ when writing an element as $r^it^j$. Essentially, there are just four cases to consider:

  • $t^j \cdot t^k$: Clearly, this is $t^{j+k}$.
  • $rt^j \cdot t^k$: Clearly, this is $rt^{j+k}$.
  • $t^j \cdot rt^k$: Note that $(t^j rt^k)(x) = (t^j r)(x+k)=t^j(-x-k)=-x-k+j$, hence $t^j \cdot rt^k = r t^{k-j}$.
  • $rt^j \cdot rt^k$: Using the previous, this is $r \cdot r t^{k-j}= t^{k-j}$.

The elements of $D_\infty$ are $t^j\colon x\mapsto x+j$ with $j\in \mathbb Z$, i.e. translations by an integer amount, and $rt^j\colon x\mapsto -x-j$, i.e. reflection around integers or half-integers (note that $-\frac j2$ is the fix pint of $x\mapsto -x-j$).

Bonus Question: Can you find any element of $G$ that is not in $D_\infty$?

Remark: How could you show that all elements have the form $r^it^j$ without observing how elemnts of $D_\infty$ are multiplied?

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Would $t^j.rt^k$ not be equal to $rt^{k-j}$ rather than $rt^{j-k}$? – HHHHHH Oct 17 '12 at 19:18
Yes as , $rt^k=t^{-k}r$ and so $t^j rt^k=t^jt^{-k}r=t^{j-k}r=rt^{k-j}$ – hmmmm Oct 17 '12 at 20:11
Thanks for the hint – Hagen von Eitzen Oct 23 '12 at 10:56

This is not an answer but a different construction of this group (which shows that it is a Coxeter group).

Let $M=\{1, -1 \}$. Then $M$ acts on $\mathbb{Z}$ by $(-1) x=-x$.

$D_{\infty}$ is the semidirect product of $M$ with $\mathbb{Z}$. The group law is given by

$$(\epsilon, x ) (\epsilon^{\prime}, x^{\prime} )=(\epsilon \epsilon^{\prime}, \epsilon^{\prime} x+ x^{\prime} ) \ . $$

If you set $\rho=(-1,0)$, $\mu=(-1,1)$, $\pi=(1,1)$ then $$\rho^2=\mu^2=1, \ \ \ \pi=\rho \mu$$

The group is generated by $<\rho, \mu>$.

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