TITLE: Elementary manipulation with elements of group
QUESTION [5 upvotes]: Let $(G,*)$ be a group with identity $e$ , let $a,b∈G$ such that $a*b^3*a^{-1}=b^2$ and $b^{-1}*a^2*b=a^3$ , then how do we prove that $a=b=e$ ?

REPLY [10 votes]: Here is a visualization of Anupam's proof. I think that this method is also applicable for other groups.
A group is just a category with one object in which each morphism is invertible. In categories we can draw commutative diagrams. The given relations can be drawn as follows:

Here $a$ represents the red vertical arrow and $b$ the green horizontal arrow. Now we reproduce the first diagram $4$ times horizontaly and then $3$ times vertically, in order to align three red arrows:

The outer rectangle can be twisted with the second relation, which yields:

But the large diagram above also contains this, but with $12$ instead of $8$ green arrows below. Thus, $4$ green arrows cancel. From parts of the large diagram we see that then $6$ green arrows cancel, hence also $2$ cancel. But then $3$ cancel, and thus the green arrow cancels. The second relation tells us that the red arrow also cancels. QED
Edit: I seem to have reinvented van Kampen diagrams.