Group with simple relations I'm working on the following algebra problem: 
Prove that if elements $a$ and $b$ of some group satisfy the relations
1) $a^5 = b^3 = 1$ and 
2) $b^{-1}ab = a^2$
then $a = 1$. 
Here is an example of what I've tried: 
If $a^5 = 1$, then using relation 2) , multiplying on the right by $a^3$ on both sides gives *$b^{-1}aba^3 = 1$. 
If $b^3 = 1$, then using relation 2), multiplying on the left by $b^4$ on both sides gives **$ab = b^4a^2$. 
Then, I tried substituting ** into *. Unfortunately, this just gets me relation 1) back again. It seems that my methods to exploit the relations always end up in simply recovering relation 1), rather than showing $a = 1$. 
Is there anything more clever I can try here? 
Thanks!  

If $b^{-1}ab=a^2$, $b^3 =1$ and $a^5=1$, then
$$\begin{align*}
b^{-2}ab^2 &= b^{-1}(b^{-1}ab)b\\
&= b^{-1}a^2b\\
&= (b^{-1}ab)^2\\
&= (a^2)^2 = a^4 = a^{-1}.
\end{align*}$$
So therefore,
$$\begin{align*}
a&= b^{-3}ab^3\\
&= b^{-1}(b^{-2}ab^2)b\\
&= b^{-1}(a^{-1})b\\
&= (b^{-1}ab)^{-1}\\
&= (a^2)^{-1}\\
&= a^3.
\end{align*}$$
Therefore, $a=a^3$.
