Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Semidirect Products (Pt. II)


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

It’s interesting then to see exactly what the semidirect product “means” in terms of subgroups. By this, I mean that the direct product of groups can be thought of as just the internal product of subgroups, but there was some cosmic accident and they actually aren’t subsets of any coherent group. The question then remains how we can interpret this for semidirect products. We begin by noticing something interesting about certain subgroups of G\rtimes_\varphi H and use this to characterize the semidirect product

Theorem: Let G and H be groups and \varphi:H\longrightarrow \text{Aut}\left(G\right). Then, if \tilde{G} and \tilde{H} are defined as before then for all (g,h)\in G\times H:

\text{ }

\begin{aligned}&\mathbf{(1)}\quad \tilde{G}\unlhd G\rtimes_\varphi H\text{ and }\left(G\rtimes_\varphi H\right)/\tilde{G}\cong H\\ &\mathbf{(2)}\quad H\leqslant G\rtimes_\varphi H\\ &\mathbf{(3)}\quad \tilde{G}\cap\tilde{H}=\{e\}\\ &\mathbf{(4)}\quad G=\tilde{G}\tilde{H}\\ &\mathbf{(5)}\quad (e,h)(g,e)(e,h)^{-1}=\left(\varphi_h(g),e\right)\end{aligned}

Proof:

\mathbf{(1)}: Define

\pi:G\rtimes_\varphi H\to H:(g,h)\mapsto h

 \text{ }

This is evidently surjective and a homomorphism. Indeed:

\text{ }

\displaystyle \pi\left((g,h)(g',h')\right)=\pi\left(\left(g\varphi_h(g'),hh'\right)\right)=hh'=\pi\left((g,h)\right)\pi\left((g',h')\right)

\text{ }

Noting though that

\text{ }

\displaystyle \ker\pi=\left\{(g,h)\in G\times H:h=e\right\}=\tilde{G}

 \text{ }

enables to conclude, from the First Isomorphism Theorem, that \tilde{G}\unlhd G\rtimes_\varphi H and \left(G\rtimes_\varphi H\right)/\tilde{G}\cong H.

\text{ }

\mathbf{(2)}: This follows from the fact that if (e,h),(e,h')\in \tilde{H} then

\text{ }

\begin{aligned}(e,h)(e,h')^{-1} &= (e,h)(\varphi_{h^{-1}}(e),h^{-1})\\ &= (e,h)(e,h'^{-1})\\ &= (e\varphi_h(e),hh'^{-1})\\ &= (e,hh'^{-1})\in \tilde{H}\end{aligned}

\text{ }

\mathbf{(3)}: This is immediate.

\text{ }

\mathbf{(4)}: Let (g,h)\in G\rtimes_\varphi H be arbitrary. Then, We merely note that

\text{ }

(g,e)(e,h)=\left(g\varphi_e(e),eh\right)=(g,h)

 \text{ }

from where the result is clear.

\text{ }

\mathbf{(5)}: This is just a simple computation:

\begin{aligned}(e,h)(g,e)(e,h)^{-1} &= (e,h)(g,e)(\varphi_{h^{-1}}(e),h^{-1})\\ &= (e,h)(g,e)(e,h^{-1})\\ &= (e,h)(g\varphi_e(e),eh^{-1})\\ &= (e,h)(g,h^{-1})\\ &= (e\varphi_{h}(g),hh^{-1})\\ &= (\varphi_h(g),e)\end{aligned}

\blacksquare

\text{ }

In fact, the first four of the above conditions characterize the semidirect product in the following sense:

\text{ }

Theorem: Let G be a group with N\unlhd G and K\leqslant G. Moreover, if N\cap K=\{e\} and G=NK then G\cong N\rtimes_\varphi K where

\text{ }

\varphi:K\longrightarrow \text{Aut}\left(N\right):h\mapsto i_h

\text{ }

where i_h is a variant of the inner automorphism i_h(n)=hnh^{-1}. (note that this makes sense that i_h\in\text{Aut}(N) since N is normal and so i_h\left(N\right)=N).

Proof: It’s evident from the conditions that G=NK and N\cap K=\{e\} that every element g of G may be writtenuniquely as the product nk with n\in N and k\in K. Indeed, by the fact that G=NK we have that every element g of G may be written as the product of n\in N and k\in K. Note though that if nk=g=n'k' then k'^{-1}k=n^{-1}n' and thus n^{-1}n'\in N\cap K so that n^{-1}n'=e or n=n' from where it quickly follows that k=k' and thus uniqueness is established. Thus, the mapping

\text{ }

\displaystyle \Phi:N\rtimes_\varphi K\to G:(n,k)\mapsto nk

\text{ }

is a bijection. Note though that

\displaystyle \begin{aligned}\Phi\left((n,k)(n',k')\right) &= \Phi\left(\left(n\varphi_k(n'),kk'\right)\right)\\ &= \Phi\left(\left(nkn'k^{-1},kk'\right)\right)\\ &= nkn'k^{-1}kk'\\ &= nkn'k'\\ &= \Phi\left((n,k)\right)\Phi\left((n',k')\right)\end{aligned}

 \text{ }

from where it follows that \Phi is a homomorphism and thus N\rtimes_\varphi K\cong G as desired. \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.


Advertisements

January 11, 2011 - Posted by | Algebra, Group Theory | , ,

1 Comment »

  1. […] Point of post: This is a continuation of  this post. […]

    Pingback by Review of Group Theory: Semidirect Products (Pt. III The Dihedral Group) « Abstract Nonsense | January 11, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: