Order of $b$ given $ab=b^{-1}a$ and $a$ of odd order. Let $G$ be a group and $a,b \in G$ such that $ab=b^{-1}a$ and
$Ord(a)$ is an odd number. Prove that $b^2=e$ where $e$ is the
identity element of $G$.
 A: Whenever possible, it is often useful to rephrase this relations in terms of conjugation. We have indeed
$$
a b a^{-1} = b^{-1}.
$$
Thus
$$
a^{2} b a^{-2} = a b^{-1} a^{-1} = (a b a^{-1})^{-1} = (b^{-1})^{-1} = b.
$$
If the order of $a$ is $2 k + 1$, we obtain
$$
b = a^{2 k + 1} b a^{- (2 k + 1)} = b^{-1},
$$
whence $b^{2} = e$.
A: Here's one way to do it, which i think is kinda fun symbolic manipulation. It's also needlessly complicated, you can do the proof with more or less the same methodology but far more concisely, try to work out an elegant proof on your own.
From your relation $ab = b^{-1}a$, you also have easily $a = bab$, $b^2a=ab^{-2}$, and $ab^2 =b^{-2}a$.  Let the order of $a$ be $2k+1$, since it's odd.  Assume $k > 0$, since $k=0$ means $a$ is the identity and it's trivial. 
Write $$
e = a^{2k+1} = (bab)^{2k+1} 
$$
In the rightmost expression, the idea is to move all the $a$s into one place by passing them through the $b^2$ groups, so that they cancel, and then to be left with only $b^2$ or $b^{-2}$.  The earlier relations tell us that when an $a$ passes through a $b^2$ barrier (or sim b^{-2}), it inverts the barrier.  
You could reason informally by parity, or if you want to see how it can easily be done formally, see below:
write $$(bab)^{2k+1}  = b(ab^2)^k a (b^2 a)^k b$$
Now consider $(ab^2)^k$.  You have $$ (ab^2)^k = (b^{-2}a)^k = b^{-2}(ab^{-2})^{k-1} a = b^{-2}(b^2a)^{k-1} a $$
These equalities just come from pulling out the end terms of the expansions or applying the relations mentioned in the beginning.  
If $k > 1$ keep going similarly to get 
$$b^{-2}(b^2a)^{k-1}  = (ab^2)^{k-2}a^2$$
This shows inductively that
$$(ab^2)^k = \begin{cases} 
b^{-2}a^k && k \equiv 1 \mod 2 \\
a^k && k \equiv 0 \mod 2
\end{cases}
$$
Symmetrically for $(b^2a)^k$. 
With this established, you can go back to the expression $e = b(ab^2)^k a (b^2 a)^k b$ and no matter the parity of $k$ you get precisely what you want.
A: In the given equation, $ab=b^{-1}a$, take the inverse of both sides to get $b^{-1}a^{-1}=a^{-1}b$; then multiply both sides of this by $a$ on the left and also on the right to get $ab^{-1}=ba$.  
This last equation says you can move a factor $a$ to the right through an adjacent $b^{-1}$ at the cost of changing that $b^{-1}$ to $b$.  The original equation says you can move a factor $a$ to the right through an adjacent $b$ at the cost of changing that $b$ to $b^{-1}$.  Combining these facts, if you have $a^nb$, for any positive integer $n$, you can move those $n$ factors $a$ to the other side of $b$, at the cost of toggling between $b$ and $b^{-1}$ each time you move one of the $a$'s.  So if $n$ is even, you get $a^nb=ba^n$, and if $n$ is odd, you get $a^nb=b^{-1}a^n$.  You're given that $a$ has odd order, so there is an odd $n$ with $a^n=e$, the identity of the group.  So, for that $n$, you get $eb=b^{-1}e$; in other "words", $b=b^{-1}$. 
