<s>[INST] If $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$ then $a=b=e$ 
Let $G$ be a group and $a,b\in  G$ such that 
  $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$. Then prove that $a=b=e$, where $e$ is the identity of group.

If we are able to prove $ab = ba$ then we are done. But i am unable to show that.
Can we use somewhere that $\gcd(2,3)=1$,
or if we write $b^2a=ab^3$ and $a^2b=ba^3$ then finding some relation between $a$ and $b$ to prove that.
 [/INST] The first equation $a^{-1}b^2a=b^3 \Rightarrow a^{-1}b^{2k}a=b^{3k}$ for all $k \in {\mathbb Z}$, so
$$a^{-2}b^{-4}a^2 = a^{-1}b^{-6}a = b^{-9}.$$
The second equation gives $a^{-2}b^{-1}a^2=ab^{-1}$, so
$$a^{-2}b^{-4}a^2 = (ab^{-1})^4$$ and hence
$$b^{-9} = (ab^{-1})^4.$$
Since $b^9$ commutes with $b$ and $(ab^{-1})^4$ commutes with $ab^{-1}$, it follows that $b^{-9} = (ab^{-1})^4 \in Z(G)$.
So $N = \langle b^9 \rangle \le Z(G)$. In the quotient $G/N$, the image of $b^4$ is conjugate to $b^9$, which is trivial, so the image of $b$ is trivial, hence the image of $a^4$ is trivial, but $a^4$ is conjugate to $a^9$, so the image of $a$ is trivial.
Hence $G/N$ has order $1$, so $G=N \le Z(G)$ and hence $G$ is abelian and it is easily proved that $G$ is trivial.
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