External semidirect product application I am trying to find all normal subgroups of $\mathbb D_n$. I've read here Normal subgroups of dihedral groups that one could show the external semidirect product $(\mathbb Z/n\mathbb Z) \rtimes (\mathbb Z/2\mathbb Z) \cong D_n$. I've just read about semidirect products and I am pretty lost on how could I define the semidirect product and then show it is isomorphic to $\mathbb D_n$. I would appreciate a proof or suggestions to show this and I would appreciate if someone could tell me how this result could be used to find all normal subgroups of the dihedral group. Thanks in advance.
 A: This is only a partial answer, because realizing $D_n$ as a semidirect product doesn't help us find all the normal subgroups in certain cases (for example, if $n$ is a power of $2$). To make the notation a little clearer, I will write $\Bbb Z_n$ and $\Bbb Z_2$ additively.
We have the homomorphism (monomorphism, actually): $\phi: \Bbb Z_2 \to \text{Aut}(\Bbb Z_n)$ given by:
$\phi(0) = \phi_0 = 1_{\Bbb Z_2}$, the identity automorphism;
$\phi(1) = \phi_1 = \text{inv}$, where $\text{inv}(x) = -x$; since $\Bbb Z_n$ is abelian, this is a homomorphism, and is clearly bijective, so an automorphism.
We can combine these with this clever trick, writing $\phi_b(x) = (-1)^bx$.
So our semidirect product binary operation becomes:
$(a,b) \ast (a',b') = (a + (-1)^ba',b+b')$ where the first coordinate is mod $n$, and the second mod $2$.
Geometrically, $(a,0)$ represents a rotation through $a/n$-ths of a circle, and $(0,1)$ represents a reflection across a line through two opposite vertices. Note that:
$(a,b) = (a,0) \ast (0,b) = (a+(-1)^00, 0+b)$, so that $(a,b)$ represents rotating, then flipping, if $b = 1$, in which case the direction of rotation is reversed.
It is easy to see that: $(1,0)$ and $(0,1)$ generate this group:
$(a,b) = (a,0) \ast (0,b) = (1,0)^a\ast(0,1)^b$; that:
$(1,0)^n = (n,0) = (0,0) = (0,2) = (0,1)^2$ and:
$(0,1)\ast(1,0) = (-1,1) = (1,0)^{-1}\ast(0,1)$ -that is we have a group of order $2n$ which satisifes the same presentation as $D_n$ ($(1,0)$ plays the role of $r$ and $(0,1)$ plays the role of $s$).
It is easy to check that:
$(1,0)\ast(a,0)\ast(-1,0) = (a,0)$ and that:
$(0,1)\ast(a,0)\ast(0,1) = (-a,0) = (a,0)^{-1}$, so any subgroup that contains only rotations is normal.
If a subgroup contains a reflection $(a,1)$ for it to be normal we must have:
$(1,0)\ast(a,1)\ast(-1,0) = (a+2,1)$ in it as well. If $n$ is odd, this implies the only normal subgroup containing a reflection is $D_n$ itself. However, if $n$ is even it only gives a necessary (but not sufficient) condition.
It turns out for even $n$ we get exactly two conjugacy classes of reflections:
$\{(a,1): a\text{ is even.}\},\ \{(a,1): a\text{ is odd.}\}$ (any normal subgroup must contain all of any given conjugacy class, so a normal subgroup that contains a reflection has at least size $\frac{1}{4}|D_n| + 1$. If $n = 2k$, our subgroup is at least of size $k + 1$. It can be shown such a subgroup must contain $\langle(2,0)\rangle$ which has $k-1$ non-identity elements), and that the only two additional normal subgroups (when $n$ is even) are:
$\langle (2,0),(0,1)\rangle$ and $\langle (2,0),(1,1)\rangle$ which are both of index $2$.
