I ran into this question when I was studying for my abstract algebra midterm.
Show that the subgroup $H$ of rotations is normal in the dihedral group $D_n$. Find the quotient group $D_n/H$.
I'm not quite sure where to begin. I know that for a Dihedral group of $n\geq 3$, then $r^n=1$ where $r$ is a rotation, and $s^2=1$ where $s$ is a reflection, and $srs=r^{-1}$. I was not sure how to prove something is a normal subgroup from here. Any advice, thanks!
On
How many rotations are there, and how does this compare to the total number of elements? You may have shown as an exercise previously that a subgroup of index $2$ is normal. That is relevant here. If not, you should prove it, because using this fact is the easiest way I see to solve your problem.
On
The index $2$ suggestion works, but you can also show this directly. One can check that the generators $R$ and $F$ of the dihedral group conform to the rule $RF = FR^{-1}$. From this, we see that any element in $D_n$ can be written as $R^jF^k$ where $0 \leq j \leq n-1$ and $0 \leq k \leq 1$.
A subgroup $N \leq G$ is normal whenever, given any $n \in N$ and $g \in G$, we have $gng^{-1} \in N$. In this case, any element of the rotation subgroup looks like $R^m$ for $1 \leq m \leq n-1$. Considering any arbitrary element $R^jF^k$ of $D_n$, we just need to show that $(R^jF^k)R^m(R^jF^k)^{-1} \in \langle R \rangle$. Clearly this is true if $k=0$, so assume $k=1$. Now look to the helpful rule in the first paragraph to conclude that this is indeed an element of $\langle R \rangle$.
If $D_n=\langle r,s\mid r^n=s^2=1,srs=r^{-1}\rangle$, then $D_n$ has order $2n$ and the group generated by $r$ has order $n$.
Therefore the index of $\langle r\rangle$ in $D_n$ is equal to two, and it is a general fact that if $H\leq G$ is a subgroup with $[G:H]=2$ then $H$ is a normal subgroup of $G$.