I would like a little bit of clarification on a basic group theory question:

True or False (Proof or Counterexample): The cosets of the subgroup of rotations R in the dihedral group $D_n$ form a cyclic group via $aR \cdot bR = abR$

My first hunch is to say false somehow because dihedral groups are not always abelian. However, the notation confuses me. From what I understand, in the context of this problem we can consider $D_3$ where "$aR$ " is the set of all "flips/rotations" of the form: (Flip or Rotation)$\cdot$(Rotation).

I'm not necessarily looking for "the answer", I just can't see where the next step might be from here.


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    $\begingroup$ What do you know about normal subgroups or the phrase "index two"? $\endgroup$ – anon Nov 14 '12 at 12:08
  • $\begingroup$ I know that index in this context is the number of cosets of R in Dn. And normal subgroups are invariant under conjugation. This hint really gave me a couple more things to think about, but unfortunately I'm having trouble linking the ideas together in the context of Dn. Any more you can give without giving the answer away? (I don't want the answer quite yet haha) $\endgroup$ – Jeff Yontz Nov 14 '12 at 12:24

$R$ is a subgroup of $D_n$ of index 2, hence is normal, hence the quotient $D_n/R$ is of order 2, hence cyclic. Note that the multiplication of cosets is the very definition of the group structure on $D_n/R$.


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