All of the dihedral groups are factor groups of the infinite dihedral group. 
Show that $\operatorname{Aut}(\Bbb{Z}) \cong \{\pm 1 \}$ and write $\alpha : \mathbb Z_2 \rightarrow \operatorname{Aut}(\Bbb{Z})$ for the nontrivial homomorphism. The semidirect product $\Bbb{Z} \rtimes_{\alpha} \Bbb{Z_2}$ is called the infinite dihedral group. Show that the dihedral groups $D_{2n}$ are all factor groups of the infinite dihedral group.

We know that $\Bbb{Z}$ only has two generators, $1$ and $-1$. So there are only two possibilities: sending $1$ to $1$ and sending $1$ to $-1$. So $\operatorname{Aut}(\Bbb{Z}) \cong \{\pm 1\}$. There is an example in our textbook that shows that $\Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z}_2 \cong D_{2n}$. So I only need to show that $$(\Bbb{Z} \rtimes_{\alpha} \Bbb{Z}_2)/\langle n \rangle \cong \Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z}_2$$ right?
Suppose that $\Bbb{Z} = \langle b \rangle$ and $\Bbb{Z}_2 = \langle a \rangle$. We can see that $(\Bbb{Z} \rtimes_{\alpha} \Bbb{Z_2})/\langle n \rangle = \{b^ia^j \text{mod n} : b^i \in \Bbb{Z}, a^j \in \Bbb{Z}_2\} = \{b^ia^j : b^i \in \Bbb{Z}_n, a^j \in \Bbb{Z}_2\} = \Bbb{Z}_n \rtimes_{\alpha} \Bbb{Z_2}$. 
Is my answer correct? I'm just curious because it seems too short. 
Thank you in advance 
 A: Your answer has a couple of issues as discussed in the comments:


*

*You are using $n$ to denote the pair $(n,0)$ (recall that the elements of $\mathbb Z\rtimes \mathbb Z_2$ are defined as pairs).

*More importantly, you are basically arguing that $\mathbb Z\rtimes \mathbb Z_2/\langle n\rangle$ and $\mathbb Z_n\rtimes \mathbb Z_2$ are equal as sets, which isn't really true or relevant; you want to show that they are isomorphic as groups. 


You are correct however in your argument that $\mathbb Z$ has only two automorphisms, and your intuition that $\mathbb Z\rtimes \mathbb Z_2/\langle (n,0)\rangle \cong \mathbb Z_n\rtimes \mathbb Z_2$ is also right. To prove this, we consider the map $f:\mathbb Z\rtimes \mathbb Z_2\to \mathbb Z_n\rtimes \mathbb Z_2$ defined by $f((a,b))=([a]_n,b)$. Clearly this is a surjective map with kernel $\langle (n,0)\rangle$. We have
$$\begin{align}
f((a,b)(c,d)) &=f((a+(-1)^bc,b+d))\\
&=([a+(-1)^bc]_n,b+d)\\
&=([a]_n+(-1)^b[c]_n,b+d)\\
&=([a]_n,b)+([c]_n,d)\\
&=f((a,b))f((c,d))
\end{align}$$
thus $f$ is a homomorphism, so by the First Isomorphism Theorem we have $\mathbb Z\rtimes \mathbb Z_2/\langle (n,0)\rangle \cong \mathbb Z_n\rtimes \mathbb Z_2$ as desired.
A: The way to show this, is to note that any of the dihedral groups have a cyclic condition.  When this condition is removed, the elements are infinite.
In essence, the cyclic groups can be written with a $mod(Z)$ condition, while the infinite group is just $z$.  Thus, every cyclic group becomes the infinite group with a cyclic remainder.
