Let $G$ be a group with identity e and distinct nonidentity elements $a$ and $b$ such that $b^3 = e$ and $(ba)^2 = e$. What is the order of $aba$? Let $G$ be a group with identity $e$ and distinct nonidentity elements $a$ and $b$ such that $b^3 = e$ and $(ba)^2 = e$. What is the order of $aba$?
This is what I did to answer this question. I want to make sure my justification is correct without any false assumptions or statements.
Knowing that $b^3 = e$ and $(ba)^2 = e$. I set 
$$b^3 = (ba)^2$$
$$b^3 = b^2*a^2$$
$$b = a^2.$$ Since we know $b^3 = e$, then we can say $a^6 = e$.
Knowing $b^3 = e$, $(ba)^2 = e$, and $a^6 = e$ we can say the order of $aba$ is equal to 6.
$$(aba)^6 = (a^6)(b^6)(a^6)= (a^6)[a^2)(b^2)]^3=(e)*(e)^3=e.$$
Thank you for the help.
 A: Starting with $$(ba)^2=baba=1$$  we multiply both sides on the left by $b^{-1}$ to get $$b^{-1}baba=b^{-1}e\implies aba = b^{-1}$$ since $b^{-1}b=e$.  
So far we have not used the assumption that $b^3=e$.
Now, the order of any element in a group is also the order of its inverse.  To see that note that if $g^n=e$ then multiplication by $g^{-n}$ yields $e=g^{-n}$.  this proves that the order of $g^{-1}$ is no greater than the order of $g$, and switching the roles of $g,g^{-1}$ yields the opposite inequality.  Thus the order of $aba$ is the order of $b^{-1}$ which is also the order of $b$, so the desired answer is $\fbox 3$.
A: Your justification is not correct. The first problem is that
$(ba)^2 = b^2 a^2$ is not true unless $a$ and $b$ commute. Remember we are not necessarily in an abelian group; $(ba)^2 = baba$ and you have to keep it in that order. Ignoring this, there is another problem, which is that you say $(aba)^6 = e$ and conclude the order of $aba$ is $6$. But it could just as easily be that the order of $aba$ is $1$, or $2$, or $3$.
To find the order of $aba$ we need to find the smallest $k$ such that $(aba)^k = e$, not just any $k$. The easiest approach, however, has already been suggested. Note that:


*

*Since $(ba)^2 = baba = e$, $aba = b^{-1}$.

*The order of $x$ and the order of $x^{-1}$ must be the same, for any $x$.
