If $a^{-1}ba=b^2$ and $|a|=3$, find $|b|$. 
Suppose $a$ and $b$ are group elements and $b\neq e$. If $a^{-1}ba=b^2$ and $|a|=3$, find $|b|$.

$b$ is also not of order $2$, otherwise $a^{-1}ba=e$, then $b=e$ contradiction.
$a^{-1}ba=b^2\Rightarrow (ab)^{3}=e$
also
$a^2ba=b^2\Rightarrow ab=a^2b^2a^2$
This leads to nowhere.
Do you know a useful manipulation ?

Squaring the given relation, we get
$$
a^{-1}b^2a = b^4
$$
We know that $b^2 = a^{-1}ba$, so inserting that, we get
$$
a^{-2}ba^2 = b^4
$$
Squaring again, and applying the same trick, gives us
$$
a^{-3}ba^3 = b^8
$$
Since $|a| = 3$, this is the same as $b^8 = b$. Therefore $b^7 = e$, which means that since $b \neq e$, we must have $|b| = 7$.
