Abstract Nonsense

Review of Group Theory: Semidirect Products (Pt. III The Dihedral Group)

Point of post: This is a continuation of  this post.

We end our very short discussion of the very interesting topic of semidirect products with a corollary( more of an example than a corollary). To the interested reader one should consult a book like Grillet (see references) to go one step further and study extension groups. Alas, time doesn’t permit me to do this as of now.

Corollary: Let $D_{n}$ be the Dihedral Group with presentation

$\text{ }$

$\displaystyle D_{n}=\left\langle r,s:r^n=e,\;\;s^2=e,\;\;srs=r^{-1}\right\rangle$

$\text{ }$

then $D_{n}\cong \mathbb{Z}_n\rtimes_\varphi\mathbb{Z}_2$ where $\varphi(1)$ is the inversion map $i(z)=z^{-1}$ (which is an automorphism since $\mathbb{Z}_n$ is abelian).

Proof: Consider $\left\langle r\right\rangle\overset{\text{def.}}{=}R$. It’s clear that $R\unlhd D_{n}$ since $\left(D_{n}:R\right)=2$ (appealing to an earlier theorem). But, it’s also clear that if $\left\langle s\right\rangle\overset{\text{def.}}{=}F$ then $F\cap R=\{e\}$. Thus, since by definition $\text{D}_{n}=RS$ we may conclude by our previous theorem that

$\text{ }$

$D_{n}\cong R\rtimes_\varphi F$

$\text{ }$

where

$\text{ }$

$\psi:F\longrightarrow R:f\mapsto i_f$

$\text{ }$

where $i_f(r)=frf^{-1}$. Note though that by definition for all $r\in R$

$\text{ }$

$i_e(r)=r$

$\text{ }$

and

$\text{ }$

$i_s(r)=srs^{-1}=r^{-1}$

$\text{ }$

So, define

$\text{ }$

$\phi:R\rtimes_\psi F\longrightarrow \mathbb{Z}_n\rtimes_\varphi\mathbb{Z}_2:\left(r^i,s^j\right)\mapsto (i,j)$

$\text{ }$

Evidently $\phi$ is a bijection and note that

$\text{ }$

$\phi\left((r^i,s^j)(r^{m},s^k)\right) = \phi\left(r^i\psi_{s^j}\left(r^m\right),s^{j+k}\right)$

$\text{ }$

but

$\text{ }$

$\displaystyle \left(r^i\psi_{s^j}\left(r^m\right),s^{j+k}\right)=\begin{cases}\left(i+m,k\right) & \mbox{if} \quad j=0 \\ (i-m,1+k) & \mbox{if} \quad j=1\end{cases}$

$\text{ }$

note though that

$\text{ }$

$\phi\left((i,j)\right)\phi\left((m,k)\right)=(i,j)(m,k)=\left(i+\psi_j(m),j+k\right)$

$\text{ }$

but

$\text{ }$

$\displaystyle \left(i+psi_j(m),j+k\right)=\begin{cases}\left(i+m,k\right) & \mbox{if}\quad j=0\\ \left(i-m,1+k\right) & \mbox{if}\quad j=1\end{cases}$

$\text{ }$

from where it follows that $\phi$ is a homomorphism, and thus an isomorphism and so

$\text{ }$

$D_{n}\cong R\rtimes_\psi F\cong \mathbb{Z}_n\rtimes_\varphi\mathbb{Z}_2$

$\text{ }$

and so the conclusion follows. $\blacksquare$

$\text{ }$

$\text{ }$

References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200

3.Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.