# Group action is faithful

Let $$G= D_{2n}$$ denote the dihedral group of order n.

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.

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.