If $a^{-1} b^2 a=b^3$ and $b^{-1}a^2 b=a^3$ with $a,b\in G$, $G$ a group, then $a=b=1$.

Question

I had this problem.

Let $G$ be a group and $a,b\in G$ such that $a^{-1} b^2 a=b^3$ and $b^{-1}a^2 b=a^3$. Prove that $a=b=1$.

Somehow I solved it, but it was a bit tedious (for example in some place I got to something like $b^3 =a^{-3}b^{-1}a^2a^{-3}b^{-1}a^2a^{-3}b^{-1}a^2=a^{-3}b^{-1}a^{-1}b^{-1}a^{-1}b^{-1}a^{2}$). I first proved that $b=a^{-2}ba^3$ (It's also true that $b=a^2ba^{-3}$) and then (Details not included) that $b^3=ab^6a^3$. Then (Details not included) that $b^3 a^3=a^{-1}b$, then $b^{-1}=a^4$ and from that $a^2=a^3$, from which $a=e$ (There are a lot of lines covering these details but there may be a mistake in these lines). It's the same to get $b=e$.

So, is there any shorter, or less tedious way to prove it?

Answer 1

This follow from Theorem $1$ and Corollary $3$ of the article Some presentations of the trivial group by Miller and Schupp, because $a^{-1}b^na=b^{n+1}$ and $a=w=b^{-1}a^2ba^{-2}$ has exponent sum $0$ in $a$.

Edit: In addition, I found the same question here.