The Dihedral group Dn is the symmetry group of the regular n-gon1.

This group has 2n elements. They are the rotations given by the powers of r, rotation anti-clockwise through 2pi/n, and the n reflections given by reflection in the line through a vertex (or the midpoint of an edge) and the centre of the polygon.

 \       |       /
   \ ----|---- /
   | \   |   / |
   |   \ | /   |
   |   / | \   |
   | /   |   \ |
   / ----|---- \
 /       |       \     
Let s denote one of these reflections.

It's not hard to see that

and that we have the relations: rn=1 s2=1 and srs-1=r-1. In fact the Dihedral group is given by these generators and relations.

If we label the vertices of the polygon by {1,2,...,n} then each symmetry of the polygon gives rise to a permutation of {1,2...,n}. This gives a homomorphism to the Symmetric group Dn->Sn. It' s easy to see that it is an injection.

In the case n=4 this is what we get.

  2          1
   |        |
   |        |
   |        |
   |        |
  3          4           
Let s be reflection in the x-axis and let r be as before. Then s corresponds to the permutation (23)(14) and t corresponds to (1234). Thus D4 is a Sylow 2-subgroup of S4.

1. There is some disagreement about notation here. Some people write D2n for what I've called Dn


If n=2k is even and n>=4, then rk is an element of order 2 which commutes with all elements of D2n. Also, rk is the only nonidentity element of D2n which commutes with all elements of D2n.

  • rk is an element of order 2 which commutes with all elements of D2n.
    rn=1 --> r2k=1
    Therefore rk is an element of order 2. |rk|=2

    To show commutability, break into two cases:
    • let ri be an element in D2n. Then (rk)(ri) = rk+i = ri+k = (ri)(rk)
    • Let ris be an element in D2n. Then (ris)(rk) = ri(srks)s = rir-ks = rk(ris).
  • Suppose some element x of D2n commutes with every other element.
    • suppose x=ri. Then xs=sx => ris = sri = r-i => ri = r-i => r2i=1, so i=k.

    Now suppose n is odd. Argue as above with x=sri and x=ri. If x=sri, then r2=1 which is not possible. If x=ri we get r2i=1 which is also not possible as then either n|2i or n|i as n is odd, so x=1. Therefore, if n is odd, only the identity commutes with all elements in D2n

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