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 $ba = ab$? 

In general, there is no middle cancellation property for groups. (If middle cancellation holds, then the group is abelian.)
Consider the group $D$ of symmetries of an equilateral triangle. For definiteness, assume the triangle has a horizontal base. The elements of this group consist of actions that leave the triangle's shape unchanged; they include:


*

*$e$, the identity;

*$r$, a rotation clockwise by 1/3 of a turn;

*$f$, a reflection left-to-right through the altitude of the triangle. 
You can check (use your mind's eye--I'm too lazy to construct the diagram) that rotation, followed by reflection, followed by rotation has the same effect as a single reflection:
$$rfr = f$$
However, two rotations don't equal the identity:
$$rr \ne e  .$$
Put another way, we have
$$ rfr = efe \quad{\rm but}\quad rr\ne ee$$
