This problem is from w.Hungerford.
My solution:
Note that $a^{-1} =a^{n-1}$ and $b^{-1}= b$.
a) $a^{2} = a a = a b b^{-1} a = a b b a = a b a^{n-1} b =a b a^{-1} b^{-1}$
So, $a^2$ is in commutator of $D_n$
Now,$D'_n= \langle a^2 \rangle$
c) for even case: n= 2m , where m is any natural number And then $a^{2m} = e $ so $ |a^2|=m$
And then $D'_n$ isomorphic to $Z_m$
But what about the odd case?
I try to prove that $|a^2|= n$, but I couldn't come up with a solution