1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.