Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. How to prove that $G$ is an abelian group? Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G$ is an Abelian group. I know that the answer for this question has been already posted and I have seen it. However, could somebody explain in detail the steps required to prove that $G$ is an Abelian group. Why, for example, could not we just divide $ababab = aaabbb$ by $ab$, get $abab = aabb$, cancel $a$ and $b$ and get $ba = ab$? 

Suppose that $a,b \in G$ are arbitary elements of the group G, with the assumption that $(ab)^3 = a^3 b^3$, and $(ab)^5 = a^5 b^5$.
Observe that for
$$
(ab)^3 = a^3 b^3
$$
Multiplying left and right by respective inverses yields
$\implies  ababab = aaabbb \implies baba = aabb $
In addition we have 
\begin{equation}
(ab)^5 = a^5 b^5 
\end{equation}
Multiplying left and right by respective inverses yeilds 
$\implies ababababab = aaaaabbbbb\implies a^{-1}abababababb^{-1}=a^{-1}aaaaabbbbbb^{-1}$
$\implies babababa=aaaabbbb$
So, now we have reduced this to a new problem. 
$(ba)^4 = a^4 b^4$ and $(ba)^2 = a^2 b^2$
Now notice that 
$(ba)^4 = (ba)^2(ba)^2 = (a^2 b^2)(a^2 b^2)$ This now shows us that the following is also true. 
$a^4 b^4 = a^2 b^2 a^2 b^2$ We now have another instance of cancellation using inverses in fact this time we can cancel twice on each side. Giving us $a^2b^2 =b^2 a^2$. We are getting close! Recall that we have the relation $(ba)^2 = a^2 b^2$. This is equivalent to saying $(ab)^2 = b^2 a^2$. All we have done is switch their roles. The rest of the proof falls out rather quickly. We now have 
$$
a^2b^2=b^2a^2= (ab)^2 = abab
$$ 
A final multiplication of inverses on each side the desired result $ab = ba$. Thus the group is Abelian. 
