Inverse of $ab$ and its order in a group with relations $a^2=e=b^6$ and $ab=b^4a$

Question

Let $a$ and $b$ be elements of a group, with $a^2=e, b^6=e$ and $ab=b^4a.$ Find the order of $ab$ and express the inverse in each of the terms $a^mb^n$ and $b^ma^n.$

Just want to cross check my solution. Would be very grateful for the complete solution.

Answer 1

From $ab=b^4a$ we get $aba^{-1}=b^4$. Taking this at both sides to the power $4$ yields $ab^4a^{-1}=b^{16}=b^4=aba^{-1}$, hence $b^4=b$, that is $b^3=e$ and $b$ has order $3$. We now see that $aba^{-1}=b^4=b$, so $a$ and $b$ commute. The order of $ab$ is therefore the l.c.m. of $3$ and $2$, which is $6$. Finally, $\langle ab \rangle$ is isomorphic to $C_6$, the cyclic group of order $6$.

Answer 2

$$ab=b^4a\\ (ab)^2=b^4aab=b^5\\ (ab)^3=abb^5=ab^6=a$$ So, indeed $(ab)^6=e$ and $(ab)^3\ne e\ne (ab)^2$.

Hence, $o(ab)=6$.

$(ab)^{-1}=(ab)^5=ab^5=b^5a$, for instance