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
Best Answer
Your answer has a couple of issues as discussed in the comments:
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.