TITLE: If $S$ is the cyclic subgroup of order n in the dihedral group $D_n$, show that $D_n/S$ is isomorphic to $Z_2$
QUESTION [6 upvotes]: If $S$ is the cyclic subgroup of order $n$ in the dihedral group $D_n$, show that $D_n/S$ is isomorphic to $Z_2$.
I know I'm supposed to find an epimorphism from $D_n$ to $Z_2$, such that $S$ is its Kernel, so that the quotient group $D_n/S$ will be isomorphic to $Z_2$. But I have no idea of how to find such an isomorphism. Besides, I don't even know what $n$ is, so I have no way of finding the elements of $D_n$ and defining an epimorphism on them.
Any help will be appreciated!!!!

REPLY [10 votes]: $D_n$ has order $2n$. If you have a subgroup $S$ of order $n$, then $D_n/S$ will have order 2 and hence is isomorphic to $\mathbb{Z}_2$.

REPLY [5 votes]: I'm not sure what you mean when you say you have no way of finding the elements of $D_n$. What definition have you been given for this group? 
$D_n$ can be defined as the group of symmetries of a regular $n$-sided polygon. It has $2n$ elements; $n$ rotations, and $n$ "flips", or reflections. The rotations form a subgroup; that's your $S$. If you can prove that it's a normal subgroup, then you know the order of the quotient group, and you know there is, up to isomorphism, only one group of that order, and you're done.