# Elementary manipulation with elements of group

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 $a=b=e$ ?

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Given $$ab^3a^{-1}=b^2$$ thus we get $$ab^6a^{-1}=b^4\;,\;\;ab^9a^{-1}=b^6$$

Replaceing $b^6$ in $ab^6a^{-1}=b^4$, we get $a^2b^9(a^{-1})^2=b^4$. Thus $a^2b^{18}(a^{-1})^2=b^8$. Again $ab^3a^{-1}=b^2$ gives $ab^{27}a^{-1}=b^{18}$. Replacing $b^{18}$, we get $a^3b^{27}(a^{-1})^3=b^8$. But $b^{-1}a^2b=a^3$. Thus $(b^{-1}a^2b)b^{27}(b^{-1}a^2b)^{-1}=b^8$. This gives $a^2b^{27}(a^2)^{-1}=b^8$. Also $a^2b^9(a^{-1})^2=b^4$ and $a^2b^{18}(a^{-1})^2=b^8$ give $a^2b^{27}(a^2)^{-1}=b^{12}$. Thus $b^8=b^{12}, i.e. b^{4}=e$. Thus $ab^6a^{-1}=e$, and thus $b^6=e$. so, $b^2=e$ and similarly it can be shown that $b^3=e$. So, $b=e$. Now showing $a=e$ is easy.

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.