# Find all conjugacy classes and normal subgroups of the Infinite dihedral group $D_\infty$

The Infinite dihedral group $D_\infty\subset \operatorname{Sym}(\mathbb{Z})$ is defined as:

$D_\infty=\{x\mapsto rx+s\mid r=\pm 1,s\in\mathbb{Z}\}$

My thoughts:

We can define the translation around $s$ as $\tau_s:\mathbb{R}\rightarrow\mathbb{R}, x\mapsto x+s$ (with $r=1$)

and the reflexion at $\frac{s}{2}$ as $\sigma_s:\mathbb{R}\rightarrow\mathbb{R}, x\mapsto s-x$ (with $r=-1$).

I can see that $D_\infty$ is generated by $\langle\tau_1,\sigma_0 \mid \sigma_0^2=e, \sigma_0\tau_1\sigma_0=\tau_1^{-1}\rangle$.

Definition from Wikipedia:

Let $G$ be a group. Two elements $a$ and $b$ of $G$ are conjugate, if there exists an element $g$ in $G$ such that $gag^{−1} = b$. One says also that $b$ is a conjugate of $a$ and that $a$ is a conjugate of $b$ .

I already found out that $\sigma_s\circ\tau_s\circ\sigma_s^{-1}=\tau_s^{-1}$. Hence, $\tau_s$ and $\tau_s^{-1}$ are in the same conjugacy class. I also tried $\tau_s\circ\sigma_s\circ\tau_s^{-1}$ and got $\sigma_{3s}$, so $\{\sigma_s,\sigma_{3s}\}$ is a conjugacy class. Also $\sigma_s$ is a conjugate of itself because of $\tau_0\circ\sigma_s\circ\tau_0^{-1}=\sigma_s$. The same is true for $\tau_s$.

My conjugacy classes so far are $\{\sigma_s\}$,$\{\tau_s\}$,$\{\tau_s,\tau_s^{-1}\}$, $\{\sigma_s,\sigma_{3s}\}$. Is this right? Which conjugacy classes am I missing?

For the second part I have a theorem which says:

Let $G$ be a group and $H\subset G$. The following are equivalent:

a) $H$ is a normal subgroup.

b) $H$ is a union of conjugacy classes of elements of G.

For example $\{\sigma_s\}\cup\{\tau_s\}= \{\sigma_s,\tau_s\}$ would be a normal subgroup of $D_\infty$ right? Does this theorem means that all possible unions of conjugacy classes are a normal subgroup?