Having trouble understanding the proof that $D_{n}$ is nonabelian for $n \geq 3$ I saw a proof that $D_{n}$ is nonabelian for $n \geq 3$ that went the following way:

Let $a,b$ are arbitrary elements of $D_{n}$. Suppose to the contrary that $ab = ba$. Then, we must have that $ab = ba \, \implies \, abb^{-1} = bab^{-1} \, \implies \, a = bab^{-1} = a^{-1}$. Therefore, $a^{2} = e$, and so $n \leq 2$.

There are two things about this proof that I do not understand:


*

*How does $a = bab^{-1}$ imply that $bab^{-1} = a^{-1}$? I have tried multiplying $ab=ba$ on both sides by $a^{-1}$s and $b^{-1}$s, and still cannot derive that $a^{-1}=bab^{-1}$. Could someone show me algebraically and in detail where this comes from?

*Also, why does $a^{2} = e$ imply that $n \leq 2$?
Could somebody please answer and explain these things to me? 
Thank you.
 A: This proof is not quite correct, but it can be fixed. Let us suppose that the definition of $D_n$ you have to work with, is that $D_n$ is the group of symmetries of a regular $n$-gon. Let $r$ denote a rotation of $2\pi/n$ about the center of  the $n$-gon, and let $s$ denote a reflection in a line passing through the center of the $n$-gon.
Then the order of $r$ is $n$, and $r$ and $s$ satisfy the following relation
$$ sr = r^{-1}s.$$
The best way to understand this relation is to try to imagine how both the left and right side act on the regular $n$-gon. To do $sr$, is to first rotate by $2\pi/n$ and reflect by $s$, and to do $r^{-1}s$ is to first reflect, and then rotate by $2\pi/n$ in the opposite direction.
We could formally prove the relation in the following way. Let the $n$-gon sit with its center at the origin of the complex plane, and $r$ be the map $z\mapsto e^{2\pi i/n}z$ and $s$ be the map $z\mapsto\overline{z}$. Its not too hard to see that $r$ does indeed represent a rotation of $2\pi/n$ radians about the origin, and the complex conjugation map $s$ represents reflection in the real axis.
Then the following computation proves the relation:
$$\overline{e^{2\pi i/n}z} = e^{-2\pi i/n} \overline{z},$$
since $r^{-1}$ is represented by multiplying by the complex number $e^{-2\pi/n}$.
Now that we know that the order of $r$ is $n$, and that $sr=r^{-1}s$, we can proceed with the proof. Suppose that $D_n$ is abelian. Then it must be true that $ab=ba$ for all $a,b\in D_n$. In particular, $rs=sr$. We already know that $sr=r^{-1}s$, hence $rs = r^{-1}s$.
Cancelling $s$ yields $r=r^{-1}$, hence $r^2=e$. Hence the order of $r$ is at most 2. However we know that the order of $r$ is $n$, hence $n\leq 2$.
It follows that $D_n$ is nonabelian when $n\geq 3$.
