Let $G= D_{2n}$ denote the dihedral group of order $2n$.
Let it act on set of all vertices $S=\{1,2,...,n\}$ of $n$-gon.
How to prove this action is faithful.
It's enough to kernel of permutation representation is just identity. Let $ \psi$ be permutation representation of action . Suppose if $g\in $ker$(\psi)$ then $g.s=s$ for all $s$ in S. Now how to prove $g=e$ ? I am just a beginner and this topic confuses me.
On
With the standard presentation of $D_{2n}$ the claim follows by just noting that $r\{1,s\}r^{-1}=\{1,r^2s\}$, and that $\{1,s\}\cap\{1,r^2s\}=\{1\}$ for $n>2$, because $r^2\ne 1$. This suffices to conclude that the action of $D_{2n}$ by left multiplication on the left quotient set $D_{2n}/H:=\{gH, g\in D_{2n}\}$ is faithful, where $H:=\{1,s\}\le D_{2n}$. In fact, the kernel of this action is: $$\bigcap_{g\in D_{2n}}gHg^{-1}\subseteq\{1,s\}\cap\{1,r^2s\}=\{1\}$$ And precisely $|D_{2n}/H|=n$.
Using the presentation $D_{2n} = \langle a, b \,|\, a^n = b^2 = 1, ab=ba^{-1}\rangle,$ we see that we can write any group element $g \in D_{2n}$ as $g=a^kb^l$ for $0 \leq k \leq n-1$, $0 \leq l \leq 1$.
You can try to explicitly compute the action of those elements on the $n$-gon to show that the only element acting trivially is the identity.