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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.

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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.

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  • $\begingroup$ Should i think a as rotation and b as reflection ? $\endgroup$ – Cloud JR Aug 8 at 13:33
  • $\begingroup$ @CloudJR Yes, exactly. $\endgroup$ – Lukas Kofler Aug 8 at 13:35

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