Register to reply

Dihedral group - isomorphism

by mehtamonica
Tags: dihedral, isomorphism
Share this thread:
mehtamonica
#1
Aug26-11, 12:34 PM
P: 26
The dihedral group Dn of order 2n has a subgroup of rotations of order n and a subgroup of order 2. Explain why Dn cannot be isomorphic to the external direct product of two such groups.

Please suggest how to go about it.

If H denotes the subgroup of rotations and G denotes the subgroup of order 2.

G = { identity, any reflection} ( because order of any reflection is 2)

I can see that order of Dn= 2n = order of external direct product
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
micromass
#2
Aug26-11, 01:44 PM
Mentor
micromass's Avatar
P: 18,334
Try to show that Dn is nonabelian (for [itex]n\geq 3[/itex]) and that your direct product will always be abelian...
mehtamonica
#3
Aug31-11, 04:24 PM
P: 26
Quote Quote by micromass View Post
Try to show that Dn is nonabelian (for [itex]n\geq 3[/itex]) and that your direct product will always be abelian...
Thanks a lot. I got it.

If we take H as the subgroup consisting of all rotations of Dn, then being a cyclic group, it would also be abelian. Then again, subgroup K of order 2 is abelian.

Further, the external direct product H + K is abelian as H and K are abelian.

Thanks !!


Register to reply

Related Discussions
Show that the matrix representation of the dihedral group D4 by M is irreducible. Calculus & Beyond Homework 1
ABSTRACT- Dihedral Group Calculus & Beyond Homework 6
Dihedral and Symmetric Group Linear & Abstract Algebra 1
Dihedral group Calculus & Beyond Homework 1
Dihedral Group of Order 8 Linear & Abstract Algebra 4