What is the center of $\displaystyle D_n$?
It seems to me that when n is odd the center is e, and when n is even there are some rotations included.
Any help is great.
i think you meant $\displaystyle \{e, a^{n/2} \}$ and the proof is not hard but requires some effort!
Edit: note that $\displaystyle D_2$ is abelian and thus it's equal to its center. so if n > 2 is even, then the center of $\displaystyle D_n$ is $\displaystyle \{e, a^{n/2} \}$.
Thanks for taking the time to answer my question. I just have a question on the notation that you used, what is $\displaystyle a^{n/2}$? does this represent a "flip" or a rotation. Also how would I begin to prove that this is abelian to every element in $\displaystyle D_n$ where n is even? Should i use a multiplication table?
recall that $\displaystyle D_n=<a,b: \ a^n=b^2=1, \ ba=a^{-1}b>=\{1,a, \cdots , a^{n-1}, b, ab, a^2b , \cdots , a^{n-1}b \}.$ since $\displaystyle a,b$ generate $\displaystyle D_n,$ an element $\displaystyle x=a^ib^j, \ 0 \leq i \leq n-1, \ 0 \leq j \leq 1,$ of $\displaystyle D_n$ is in the center of
$\displaystyle D_n$ if and only if $\displaystyle ax=xa$ and $\displaystyle bx=xb.$ this will simplify your job a lot!