I am trying to calculate the commutator of the Dihedral group. If $n=1,2$ then $[D_n,D_n]=1$. Now I consider the case $n\geq 3$.
I thought of using the property $[G,G] \subset H$ iff $H \lhd G, G/H$ is abelian. So now my problem reduces to finding all normal subgroups of $D_{2n}$ and look which one satisfies $G/H$ is abelian. I've proved that $H=\langle R \rangle$ is normal. On the other hand, we have $$(1) \space (r^is^jH)(r^ks^lH)=(r^is^jr^ks^l)H$$$$=(r^ir^{-k}s^js^l)H$$
$$(2) \space (r^ks^lH)(r^is^jH)=(r^ks^lr^is^j)H$$$$=(r^kr^{-i}s^ls^j)H$$
So (1)=(2) iff $(r^ir^{-k}s^js^l){(r^kr^{-i}s^ls^j)}^{-1} \in H$.
But $(r^ir^{-k}s^js^l){(r^kr^{-i}s^ls^j)}^{-1}=(r^ir^{-k}s^js^l)(s^js^lr^{i}r^{-k})$$$=r^ir^{-k}r^ir^{-k} \in H$$
Note that since $D_{2n}$ is not abelian, then $[D_{2n},D_{2n}]\neq 1$.So $[D_{2n},D_{2n}] \subset \langle R\rangle$. If $n=p$ with $p$ prime, then $[D_{2n},D_{2n}]=\langle R \rangle$.
I have no idea what to do for the case $n$ composite number, I would really appreciate some help to analyze this case.
Since $srsr^{-1}=r^{-2}$, every even power of $r$ is in the commutator. So if $n$ is odd, that's every power of $r$, and since $\langle r\rangle$ is normal and has abelian quotient ($C_2)$, this is exactly the commutator subgroup.
If $n$ is even, then the commutator is $\langle r^2\rangle$, since its quotient is $V_4$, which is abelian.