# If S is the cyclic subgroup of order n in the dihedral group Dn, show that Dn/S is isomorphic to Z2

If S is the cyclic subgroup of order n in the dihedral group Dn, show that Dn/S is isomorphic to Z2.

I know I'm supposed to find an epimorphism from Dn to Z2, such that S is its Kernel, so that the quotient group Dn/S will be isomprphic to Z2. 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 Dn and defining an epimorphism on them.

Any help will be appreciated!!!!

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$D_n$ 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 $\mathbb{Z}_2$.

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Thanks!!!I haven't thought of that... I've been thinking about finding an isomorphism –  Scharfschütze Nov 3 '11 at 2:14
It should perhaps be noted that subgroups of index 2 are always normal, so the quotient group exists. –  Gerry Myerson Nov 3 '11 at 2:41
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.