Problem: Show that there exists a monomorphism from $D_n$ to $S_n$, $n\geq 3$.
Write $D_n=\langle x,y\mid x^n=1, y^2=1, yx=x^{-1}y\rangle$.
Define a map $\phi:D_n\to S_n$ such that $ \phi(x)=\begin{pmatrix}1&2&...&n\\2&3&..&n\end{pmatrix}$ and
$\phi(y)=\begin{pmatrix}1&2&3&...&n-1&n\\1&n&n-1&...&3&2\end{pmatrix}$.
I want to show that $\phi$ is a homomorphism. Let $a\in D_n$. Then $a=x^jy^k$ for some $j,k\in\mathbb{Z}$. So we need :$\phi(x^jy^k)=\phi(x^j)\phi(y^k)$ (I don't know how to do that?).
There is only such homomorphism:
Suppose there is other homomorohism say $\psi:D_n\to S_n$ such that $\psi(x)=\phi(x)$ and $\psi(y)=\phi(y)$.
Let $a\in D_n$. Then $a=x^jy^k$ for some $j,k\in\mathbb{Z}$.
Then $\phi(x^jy^k)=\phi(x^j)\phi(y^k)=(\phi(x))^j(\phi(y))^k=(\psi(x))^j(\psi(y))^k=\psi(x^j)\psi(y^k)=\psi(x^jy^k)$.
Thus $\phi=\psi$. We're done(Is it correct?).
Call $H:=\{1,y\}\le D_n$. Now consider the action of $D_n$ by left multiplication of the left quotient set $D_n/H$. This action has kernel $\bigcap_{g\in D_n}g^{-1}Hg$, which is trivial for $n>2$ because then $H\cap x^{-1}Hx$ is trivial: in fact, $x^{-1}Hx=\{1,x^{-1}yx\}=\{1,yx^2\}$, and $y\ne yx^2$ as soon as $n>2$ (because then $x^2\ne 1$). But the size of $D_n/H$ is precisely $n$, and hence $D_n$ acts faithfully on a set of $n$ elements or, equivalently, it embeds into $S_n$.