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Dihedral group - isomorphism

by mehtamonica
Tags: dihedral, isomorphism
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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
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micromass
#2
Aug26-11, 01:44 PM
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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 !!


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