# 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.

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What do you mean by $\langle n\rangle$? Do you mean the subgroup generated by $n$-th powers? Or are you identifying $n\mathbf{Z}$ with a subgroup of the semi-direct product? Showing that the quotient modulo some particular subgroup is isomorphic $D_{2n}$ would be enough, but strictly speaking there's no reason that you must use that subgroup. Thinking in terms of the first isomorphism theorem, you don't need to identify a particular subgroup with quotient isomorphic to $D_{2n}$. You only need to write down a surjective homomorphism to $D_{2n}$. – Keenan Kidwell May 18 '13 at 22:35
Also keep in mind that your group is non-abelian, so the set of $n$-th powers of elements of the group might not form a subgroup. You also have to worry about normality. I think in this case it is much easier to write down a surjective homomorphism to $D_{2n}$. – Keenan Kidwell May 18 '13 at 22:37
Dear @Artus, I assume by $n$ you actually mean $(n,1)$ (i.e. you are identifying $\mathbf{Z}$ with its natural image in the semi-direct product). What you say about orders of elements in the image is false. It is true for elements of the form $f(x)$ where $x$ has finite order. If $x$ has infinite order there is no reason that $f(x)$ must have finite order. Saying that one group $H$ is a factor of another group $G$, in this context, means that $H$ is a homomorphic image of $G$, i.e., that there is a surjective homomorphism $G\rightarrow H$. Incidentally, I think "factor" is not so good a word – Keenan Kidwell May 18 '13 at 23:15
Agree with Keenan's remarks. And about that the OP probably asks whether the finite dihedral groups are quotient groups of the infinite one. But the basic idea in your solution is sound. – Jyrki Lahtonen May 20 '13 at 7:07
But I would also approach the problem differently, using definitions of these groups in terms of their presentations. The infinite dihedral group is generated by the elements $r$ and $s$ subject to the relations $s^2=1, srs=r^{-1}$. The finite one have the additional relation $r^n=1$ for some $n$. How could we get a surjective homomorphism from one to the other?? Hmm. Some call this von Dyck's Theorem, but I am used to thinking about it as a rather straight forward consequence of the idea of group presentations. To each their own! – Jyrki Lahtonen May 20 '13 at 7:11

• 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.
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.