Prove that the Dihedral group $D_n$ is isomorphic to $Z_n \rtimes_{\psi} Z_2$ I consider the following map $\psi : Z_2 \rightarrow Aut(Z_n)$ where we map 
the identity element 0 to the identity map and $1 \mapsto \theta : Z_n \rightarrow Z_n$ where $\theta(x) = -x$. I am not sure how to proceed further. Any help would be nice.
$D_n = ⟨ r, s : r^n = e = s^2, r^js = sr^{-j} ⟩ $
 A: Note that any element $\sigma\in D_n$ can be written uniquely in the form $\sigma=r^is^j$ for some $0\leq i<n$ and $j=0$ or $1$. Thus can be seen using the relation $r^{i}s=sr^{-i}$, which allows us to move all of the occurrences of $r$ in a word to the left and all occurrences of $s$ to the right. As $|D_n|=2n$, we see that, in fact, the map $\phi:Z_n\times Z_2\to D_n$ defined by
$$
\phi((i,j))=r^is^j
$$
is a bijection. This $\phi$ is not only a bijection of sets, but is a homomorphism once the set $Z_n\times Z_2$ is endowed with the group structure $Z_n\rtimes Z_2$. This is because
$$
\phi((i_1,j_1))\phi((i_2,j_2))=r^{i_1}s^{j_1}r^{i_2}s^{j_2}=r^{i_1+\psi(j_1)(i_2)}s^{j_1+j_2}=\phi((i_1+\psi(j_1)(i_2),j_1+j_2)),
$$
which is precisely $\phi((i_1,j_1)(i_2,j_2))$. So $\phi$ gives us an isomorphism between $D_n$ and $Z_n\rtimes Z_2$.
The $\psi(j_1)(i_2)$ term appears in the exponent of the $r$ in the second equality above because if $j_1=0$, then $r^{i_1}s^{j_1}r^{i_2}s^{j_2}=r^{i_1+i_2}s^{j_2}$, and if $j_1=1$, then $r^{i_1}s^{j_1}r^{i_2}s^{j_2}=r^{i_1-i_2}s^{1+j_2}$, using the relation $sr^{i}=r^{-i}s$
A: Perfectly good answer was already given by Galena Rupp, but I will try to give intuition why the given isomorphism really is a natural choice.
First of all, let us start with general semidirect product $N\rtimes_\psi H$. Instead of ordered pair $(n,h)$, I will simply write $nh$ (where concatenation is multiplication in $N\rtimes_\psi H$). So, how do we multiply $(n_1h_1)(n_2h_2)$? Well, $(n_1h_1)(n_2h_2) = n_1(h_1n_2h_1^{-1})h_1h_2 = (n_1\psi_{h_1}(n_2))(h_1h_2)$ which is precisely how we define multiplication in semidirect product. This tells us that $\psi$ is encoding how $H$ acts by conjugation on $N$ in the semidirect product - with this information we can completely determine the structure of semidirect product.
Let us get back at $\Bbb Z_n\rtimes_\psi \Bbb Z_2$. First I will switch to multiplicative notation $C_n\rtimes_\psi C_2$, denote the generator of $C_n$ with $r$ and the generator of $C_2$ with $s$, and appropriately change $\psi_s$ to $r^j\mapsto r^{-j}$. Now, if I want to give presentation of $C_n\rtimes_\psi C_2$, I note that it is generated by $r,s$ subject to conditions $r^n = e$, $s^2 = e$ and $\psi_s(r^j) = r^{-j}$, i.e. $$C_n\rtimes_\psi C_2 = \langle r,s\,|\, r^n = s^2 = e, \psi_s(r^j) = r^{-j}\rangle$$
Now, remember that $\psi_s$ really is a conjugation, so rewrite the above as $$C_n\rtimes_\psi C_2 = \langle r,s\,|\, r^n = s^2 = e, sr^js^{-1} = r^{-j}\rangle = \langle r,s\,|\, r^n = s^2 = e, r^js = sr^{-j}\rangle$$ which is precisely the presentation of $D_n$.
So, what is the isomorphism $\Bbb Z_n\rtimes_\psi \Bbb Z_2\cong D_n$? Well, from the above, we just have to rename generators to $r$ and $s$ and switch to multiplicative notation, i.e. the isomorphism is given by $(i,j)\mapsto r^is^j$ as noted in the answer by Galena Rupp.
