Why can't Dn be isomorphic to the direct product of its subgroups?

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SUMMARY

The dihedral group Dn of order 2n cannot be isomorphic to the external direct product of its subgroups, specifically the subgroup of rotations (H) of order n and the subgroup of order 2 (G). This is established by demonstrating that Dn is nonabelian for n ≥ 3, while the external direct product H × G is always abelian. Since both H and G are abelian, their direct product cannot capture the nonabelian structure of Dn, confirming the impossibility of such an isomorphism.

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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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Try to show that Dn is nonabelian (for n\geq 3) and that your direct product will always be abelian...
 
micromass said:
Try to show that Dn is nonabelian (for n\geq 3) 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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