Assuming $ab^2 = b^3a$ and $a^2=1$ prove that the order of $b$ is $5$. Let $G$ be a group and $a,b \in G$ with $a \ne 1$ and $b \ne 1$. Assuming $ab^2 = b^3a$ and $a^2=1$ I need to prove that the order of $b$ is $5$. 
I have proved by contradiction that it can't be 2 or 3 but I don't know how to prove that it must be 5 and it can't be 4.

Note that 
$$b^5 = b^3a^2b^2=(b^3a)(ab^2)=(b^3a)(b^3a)=b^3(ab^2)ba=b^3(b^3a)ba=b^6aba,$$
multiplying by $b^{-5}$ from the left we get $1=baba$. Multiplying by $a$ from the right we get $a=baba^2=bab$. Thus 
$$1=a^2=(bab)^2=b(ab^2)ab=b(b^3a)ab=b^4a^2b=b^4\cdot 1 \cdot b = b^5.$$
