**COMMUTABILITY OF ELEMENTS IN D**

_{2n}If n=2k is even and n>=4, then r^{k} is an element of order 2 which commutes with all elements of D_{2n}. Also, r^{k} is the only nonidentity element of D_{2n} which commutes with all elements of D_{2n}.

__PROOF__:

- r
^{k}is an element of order 2 which commutes with all elements of D_{2n}.

r^{n}=1 --> r^{2k}=1

Therefore r^{k}is an element of order 2. |r^{k}|=2

To show commutability, break into two cases:- let r
^{i}be an element in D_{2n}. Then (r^{k})(r^{i}) = r^{k+i}= r^{i+k}= (r^{i})(r^{k}) - Let r
^{i}s be an element in D_{2n}. Then (r^{i}s)(r^{k}) = r^{i}(sr^{k}s)s = r^{i}r^{-k}s = r^{k}(r^{i}s).

- let r

_{2n}commutes with every other element.

- suppose x=r
^{i}. Then xs=sx => r^{i}s = sr^{i}= r^{-i}=> r^{i}= r^{-i}=> r^{2i}=1, so i=k.

Now suppose n is odd. Argue as above with x=sr^{i} and x=r^{i}. If x=sr^{i}, then r^{2}=1 which is not possible. If x=r^{i} we get r^{2i}=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 D_{2n}