- Show that if \(N\normal D_n\) is a normal subgroup of a dihedral group, then \(D_n/N\) is again a dihedral group.
Solution
The literal claim fails for \(N=D_n\), whose quotient is trivial. Every other quotient is dihedral, where \(D_1\cong C_2\) and \(D_2\cong C_2\times C_2\) are the dihedral groups of orders \(2\) and \(4\).
Write \[ D_n=\langle r,s\mid r^n=1,\ s^2=1,\ srs^{-1}=r^{-1}\rangle, \] let \(Q=D_n/N\), let \(\bar r=rN\) and \(\bar s=sN\), and let \(k\) be the order of \(\bar r\), so \(k\mid n\).
1. \(Q\) is generated by \(\bar r\) and \(\bar s\), and\[ \bar r^k=1, \qquad \bar s^2=1, \qquad \bar s\bar r\bar s^{-1}=\bar r^{-1}. \]
The quotient map is a surjective homomorphism, so it carries the generators \(r,s\) to generators of \(Q\) and the defining relations of \(D_n\) to relations in \(Q\); \(\bar r^k=1\) is the definition of \(k\).
2. Every element of \(Q\) has the form \(\bar r^i\) or \(\bar s\bar r^i\) with \(0\le i<k\).
The relation \(\bar r\bar s=\bar s\bar r^{-1}\) moves every occurrence of \(\bar s\) in a word to the left, \(\bar s^2=1\) leaves at most one occurrence, and powers of \(\bar r\) reduce modulo \(k\).
3. If \(\bar s\notin\langle\bar r\rangle\), then \(Q\cong D_k\).
The elements \(\bar r^i\) with \(0\le i<k\) are distinct because \(\bar r\) has order \(k\), and so are the elements \(\bar s\bar r^i\). If \(\bar s\bar r^i=\bar r^j\), then \(\bar s=\bar r^{j-i}\in\langle\bar r\rangle\), contrary to hypothesis. So the \(2k\) elements of 2 are distinct and \(|Q|=2k\). By 1, \(r\mapsto\bar r\), \(s\mapsto\bar s\) defines a surjective homomorphism \(D_k\to Q\), which is an isomorphism because both groups have order \(2k\).
4. If \(\bar s\in\langle\bar r\rangle\), then \(Q\) is cyclic of order \(1\) or \(2\).
By 1, \(Q=\langle\bar r\rangle\), so \(Q\) is abelian and \[ \bar r=\bar s\bar r\bar s^{-1}=\bar r^{-1}. \] Hence \(\bar r^2=1\) and \(|Q|=k\le2\).
5. \(Q\) is dihedral unless \(N=D_n\).
3.Otherwise 4 gives \(|Q|\le2\); if \(|Q|=2\) then \(Q\cong C_2\cong D_1\), and \(|Q|=1\) means \(N=D_n\).