Let $G$ be a group and $a,b \in G$ show that $|b| = 5$ 
Show that if $|a|=2$, $ab^2a^{-1}=b^3$ and $b\ne e$, then $|b|=5$.

My attempt
$$a^2 = e \implies a * a = e \implies a = a^{-1}$$
$$ab^2a=b^3$$
$$(ab^2a)^2=(b^3)^2$$
I) $$ab^2aab^2a = b^5$$
$$ab^2a=b^3$$
$$ab^2ab^2=b^3b^2$$
II) $$ab^2ab^2 = b^5$$
I and II$\implies$$b^4 = b^5 \implies b = e \implies b^5 = e$
Of course it is wrong, I've been trying to solve this problem for over 2 hours but I can't find a way to show $b^5=e$ through the given properties.

One has
$ab^2 = b^3a$, then $(ab^2)^2 = b^3ab^3a = b^3a(ab^3a)ba = b^6aba$
$b^5 = b^3b^2 = ab^2a^{-1}b^2=ab^2ab^2=(ab^2)^2 = b^6aba$.
So, $baba = e$, then $b^{-1} = aba$, or $b^{-2} = ab^2a = b^3$, then $b^5=e$. 
