Does $D_{15}$, the dihedral group of order 30 have a normal subgroup of order 6? This question is from Artin Algebra: List the proper normal subgroups N of the dihedral group $D_{15}$. I have most of the answer, partly from Takumi Murayama's solutions to Artin: https://www.csie.ntu.edu.tw/~b01902113/artin-sols.pdf
Let $D_{15}= \{1,x, ..., x^{14}, y, yx, ... ,yx^{14}\}$, with the relations  $$x^{15} = 1, y^{2} = 1, yx = x^{14}y$$
The orders of the proper subgroups can be $2, 3, 5, 6, 15$. Every element of the form $x^{i}y$ generates an order $2$ subgroup, because $$x^{i}y*x^{i}y = x^{i}x^{14i}yy = x^{15i} = 1$$ These subgroups of order $2$ can't be normal, because conjugating $x^{i}$ by $x^{14}$, $$x^{14}x^{i}yx = x^{14 +i}yx = x^{28+i}y = x^{13+i}y \neq x^{i}y$$
Thus, the subgroups H of orders $3, 5, 15,$ can't contain an element $x^{i}y$, because if they did, we would have $\langle x^{i}y\rangle \subseteq H$ which is not possible because $2 \nmid 3, 5, 15$. Hence, the subgroups of orders $3, 5, 15,$ are the cyclic, normal, subgroups $\langle x \rangle, \langle x^3
\rangle, \langle x^{5} \rangle$. 
My question is what about the subgroups of order $6$; $2$ does divide $6$, so the above reasoning does not work. Is $6$ a normal subgroup or not? why?
 A: A subgroup of order 6 cannot be cyclic, as $D_{15}$ has no elements of order 6. Thus, it must be isomorphic to $C_2 \times C_3$, so must be generated by some elements $a,b$, of orders $2$ and $3$ respectively. 
Now, the elements of $D_{15}$ of order $3$ are precisely $x^5$ and $x^{10}$ (so both must be in our subgroup), while the elements of order $2$ are precisely the $x^ay$. Thus, our subgroup must be of the form $G_a = \{1,x^{5}, x^{10}, x^ay,x^{5+a}y, x^{10+a}y\}$ for some integer $a$.
Now, we simply note that $xG_ax^{-1}$ contains $x^{a+1}yx^{-1} = x^{a+2}y$, which does not lie in $G_a$, hence $xG_ax^{-1}\neq G_a$, so $G_a$ is not normal in $D_{15}$.  
A: There is a full classification of all normal subgroups of dihedral groups. Just check the case $n=15$.
Normal subgroups of dihedral groups
More precisely: For $n$ odd, the normal subgroups are given by $D_n$ and $\langle R^d \rangle$ for all divisors $d\mid n$. The groups $\langle R^d \rangle$ have index $2d$, so index $5$ for $n=15$ is impossible. So there is no normal subgroup of order
$\frac{30}{5}=6$.
