Let, $G$ be a group and $a,bin G$. If $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$ then show that $a=b=e$, where $e$ is the identity of $G$.

We have from the first condition, $b^{27}=a^{-1}b^{18}a$. Then what to do?

Any hint.?

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# A basic question on group theory

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Let, $G$ be a group and $a,bin G$. If $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$ then show that $a=b=e$, where $e$ is the identity of $G$.

We have from the first condition, $b^{27}=a^{-1}b^{18}a$. Then what to do?

Any hint.?

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