Show that if \(N\trianglelefteq D_n\) is a normal subgroup of a dihedral group, then \(D_n/N\) is again a dihedral group.
Solution
Write \[ D_n=\langle r,s\mid r^n=s^2=1,\ srs=r^{-1}\rangle, \] and let \(N\trianglelefteq D_n\). Put \(\bar r=rN\) and \(\bar s=sN\). Then \(D_n/N\) is generated by \(\bar r,\bar s\) and satisfies \[ \bar s^2=1, \qquad \bar s\bar r\bar s^{-1}=\bar r^{-1}. \] Let \(d=|\bar r|\), so \(d\mid n\).
If \(\bar s\notin\langle\bar r\rangle\), then every element of the quotient has a unique form \(\bar r^i\) or \(\bar s\bar r^i\). Thus the quotient has order \(2d\). If \(d\ge2\), it is the usual dihedral group \[ D_n/N\cong D_d. \] If \(d=1\), then \(\bar r=1\) and \(D_n/N=\langle\bar s\rangle\cong C_2\).
If \(\bar s\in\langle\bar r\rangle\), then the quotient is cyclic. But the relation \[ \bar s\bar r\bar s^{-1}=\bar r^{-1} \] becomes \(\bar r=\bar r^{-1}\), hence \(\bar r^2=1\). Therefore the quotient is either trivial or cyclic of order \(2\).
Thus every quotient is either an ordinary dihedral group \(D_d\) with \(d\ge2\), or one of the two degenerate cases \(1\) and \(C_2\). If one adopts the conventions \(D_0=1\) and \(D_1=C_2\), then every quotient of \(D_n\) is again dihedral.