# Infinite Dihedral group

• Jan 8th 2009, 10:31 PM
bulls6x
Infinite Dihedral group
Here's a problem I had last semester in my grad Algebra class which really helped me understand short exact sequences that split.

Let $D_{\infty}$ be a group with presentation $$

Show that $G \simeq\mathbb{Z}\rtimes\mathbb{Z}/2$.

Hint: Find a short exact sequence which splits. The sequence should be clear from the isomorphism.
• Jan 9th 2009, 09:21 AM
NonCommAlg
Quote:

Originally Posted by bulls6x
Here's a problem I had last semester in my grad Algebra class which really helped me understand short exact sequences that split.

Let $D_{\infty}$ be a group with presentation $$

Show that $G \simeq\mathbb{Z}\rtimes\mathbb{Z}/2$.

Hint: Find a short exact sequence which splits. The sequence should be clear from the isomorphism.

no need for Hint! (Smirk) the question is straightforward: every element of $D_{\infty}$ is in the form $(ab)^na^k,$ where $n \in \mathbb{Z}, \ k \in \{0,1\}.$ now define $f: \mathbb{Z}/2 \longrightarrow D_{\infty}$ by $f([k])=a^k,$ and

$g: D_{\infty} \longrightarrow \mathbb{Z}$ by $g((ab)^n a^k)=n.$ see that $f,g$ are homomorphisms and $\ker g=\text{im} f.$ also $h: \mathbb{Z} \longrightarrow D_{\infty}$ defined by $h(n)=(ab)^n$ satisfies $gh=\text{id}_{\mathbb{Z}}.$ therefore the sequence

$1 \longrightarrow \mathbb{Z}/2 \longrightarrow D_{\infty} \longrightarrow \mathbb{Z} \longrightarrow 1$ is exact, split and hence $D_{\infty} \simeq \mathbb{Z} \rtimes \mathbb{Z}/2.$