Dihederal Group $D_{2n}$ Where $n$ is even/odd

Question

I know that the group presentation of $D_{2n}$ is the following

$$D_{2n} = \big<a,b: a^n=b^2=1,b^{-1}ab =a^{-1} \big>$$

Now if we consider the case where $n$ is even and we write $n =2m$ for some integer $m$. Why is it true that $b^{-1}a^mb = a^{-m} = a^m$

I know that $b^{-1}a^mb = a^{-m}$ follows from the fact that $b^{-1}ab = a^{-1}$ and so $a^{-2} = a^{-1}a^{-1} = (b^{-1}ab)(b^{-1}ab) = b^{-1}a^2b$ and so in general we do have that $b^{-1}a^kb =a^{-k}$ for any integer $k$ and so in particular we do have that $b^{-1}a^mb=a^{-m}$.

However, How did we get the equality that $a^m=a^{-m}$ in this case ?

Another question, if $n$ is odd. Then is it true that $a^{-i} \neq a^i$ for any integer $i$. I guess it is true because if $a^{-i} = a^i$ then this will imply that $a^{2i} = 1$, However, $2i$ is even and we already assumed that $n$ is odd and so we obtain a contradiction (Is that right/rigorous) ?

Answer 1

For your first question: Note that $a^ma^m=a^{2m}=1$, so $a^{-m}=(a^m)^{-1}=a^m$.

For your second question: You are on the right track, but your proof is not entirely rigorous. What about $i=0$? Or $i=n$?