# Math Help - Dihedral Groups

1. ## Dihedral Groups

What is the center of $D_n$?

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.

2. you're right. if n is odd, the center is e.
now, if n is even, the center is $\{e, a^{n/2}\}$. this should be easy to prove.

3. Originally Posted by kalagota

now, if n is even, the center is $\{e, a^{n/2}b\}$. this should be easy to prove.
i think you meant $\{e, a^{n/2} \}$ and the proof is not hard but requires some effort!

Edit: note that $D_2$ is abelian and thus it's equal to its center. so if n > 2 is even, then the center of $D_n$ is $\{e, a^{n/2} \}$.

4. Thanks for taking the time to answer my question. I just have a question on the notation that you used, what is $a^{n/2}$? does this represent a "flip" or a rotation. Also how would I begin to prove that this is abelian to every element in $D_n$ where n is even? Should i use a multiplication table?

5. Originally Posted by Order 2
Thanks for taking the time to answer my question. I just have a question on the notation that you used, what is $a^{n/2}$? does this represent a "flip" or a rotation. Also how would I begin to prove that this is abelian to every element in $D_n$ where n is even? Should i use a multiplication table?
recall that $D_n==\{1,a, \cdots , a^{n-1}, b, ab, a^2b , \cdots , a^{n-1}b \}.$ since $a,b$ generate $D_n,$ an element $x=a^ib^j, \ 0 \leq i \leq n-1, \ 0 \leq j \leq 1,$ of $D_n$ is in the center of

$D_n$ if and only if $ax=xa$ and $bx=xb.$ this will simplify your job a lot!

6. Originally Posted by NonCommAlg
i think you meant $\{e, a^{n/2} \}$ and the proof is not hard but requires some effort!
yeah, that's what i meant. tnx! i edited it already!