Dihedral group is supersolvable

Question

I need to show that Dihedral group $D_n$ is supersolvable.

My Approach : I think the existence of a normal chain $\{e\} = G_0 \leqslant G_1 \leqslant ... \leqslant G_n = G$ satisfying following

1. $G_i \unlhd G$ $\forall i$

2. $G_i / G_{i-1}$ cyclic

So I claim that ${e} \subset \left<r\right> \subset D_n$ satisfies the conditions for supersolvable ..

so I denote $C_n = \left<r\right>$ a cyclic group generated by $r$ which is a subgroup of $D_n$

1. Need to show that $C_n \unlhd D_n$ so let $g \in D_n$ so $g=s^kr^m$ where $k=0 $ or $k=1$. Now $gr^lg^{-1} = (s^kr^m)r^l (s^kr^m)^{-1}$ here $r^l \in C_n$ thus $gr^lg^{-1} = (s^kr^m)r^lr^{-m}s^{-k} = s^kr^ls^{-k}$. Now if $k=0$ then $gr^lg^{-1} = r^l$ thus proved to be normal else for $k=1$ then $gr^lg^{-1} = sr^ls^{-1} = r^{-l} \in C_n$ thus proved to be normal.

Now I am stuck with how to show that $D_n / C_n $ is cyclic Also not sure if the above part s correct.. Please help

Answer 1

The rotation subgroup $C_n$ has index $2$ in $D_n$, and index $2$ subgroups are always normal. Also the classification of groups of order $2$ is very simple: there is up to isomorphism one such group, and it is cyclic. (The latter part is true for any prime order, but for order $2$ it is particularly basic: one can just consider the possibilities for the multiplication table, and quickly see there is only one possiblilty.)

(Many proofs exist, but most elementarily: every subgroup has at least one coset that is both a left and right coset, namely that of the identity, which coset is the subgroup itself. But for an index $2$ subgroup this leaves just one other coset, its complement, which then is both left coset and right coset as well.)