For a $n$-sided regular polygon there are $n-1$ possible rotations: $a,a^2,a^3,a^{n-1}$, a 1 reflection $b$, 1 identity $e=a^n=b^2$. There are also 2(n-1) elements $ab,a^2b,a^3b,…a^{n-1}b$ and $ba,ba^2,ba^3,…ba^{n-1}$ (since rotation and reflections do not commute). Therefore, a dihedral group has $(n-1)+1+1+2(n-1)=3n-1$ elements!
Where is my counting wrong?
Best Answer
It is true that $a$ and $b$ do not commute, but that does not mean that the elements $ba,ba^2,ba^3,...ba^{n-1}$ are all distinct from the elements $ab,a^2b,a^3b,...a^{n-1}b$. Indeed, with the usual choice of generators for the dihedral group, we have $ba=a^{n-1}b$, and it follows that $ba^r=a^{n-r}b$ for all $r$ and so your last $n-1$ elements are all repeats of elements you already had.