Quotient Group $D_2n/R$ is abelian where $R$ is the group of rotations? $D_2n$ is not abelian. However, the group of rotations, denoted $R$, is. I've already shown that $R$ is a normal subgroup of $D_2n$; however I'm stuck at showing the quotient group is abelian.
I know if it is abelian, $xRyR=yRxR$ but I get stuck at $xRyR=(xy)R$.
But $x$ and $y$ are not necessarily commutative. How do I continue? 
 A: Since $R=C_n$ is a subgroup of $D_{2n}$ of index $2$, it is a normal subgroup. The quotient $D_{2n}/C_n$ has exactly $2$ elements, because of
$$
|D_{2n}/C_n|=\frac{|D_{2n}|}{|C_n|}=\frac{2n}{n}=2.
$$
 But every group of order $2$ is abelian (there is only one).
A: Dietrich's proof is fine, but I wanted to show you how you could finish off your own proof.
First note that $R$ has order $n$ and $D_{2n}$ has order $2n$, so $D_{2n}/R$ consists of two elements.  Of course, one is the coset containing the identity, $1R$, and the other is given by taking any reflection $y$ and creating the coset $yR$.
We need to check that $(1R)(1R) = (1R)(1R)$, that $(yR)(yR) = (yR)(yR)$, and that $(1R)(yR) = (yR)(1R)$.  The first two follow automatically.
The last follows because the identity commutes with everything: we have $$(1R)(yR) = (1y)R = (y1)R = (yR)(1R).$$
One last ponit:  we never explicitly used the fact that $y$ is a reflection, just that $y\notin R$.  So this proof applies whenever we have an index two subgroup (which is then automatically normal).  This gives another way of looking at Dietrich's proof, which only needed that fact that $|D_{2n}/R| = 2$.
