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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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    $\begingroup$ Nice question...darun @ Souvik $\endgroup$
    – Anupam
    Oct 2, 2013 at 14:15

2 Answers 2

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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.

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    $\begingroup$ Oh, dear! Please do use more generously spacing, in particular much more lines spacing! Note that if you use double dollar signs to open and close mathematical expressions, these are written in one single, different line. I did the first ones for you. $\endgroup$
    – DonAntonio
    Oct 2, 2013 at 15:31
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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:

enter image description here

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:

enter image description here

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

enter image description here

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.

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  • $\begingroup$ Very nice explanation and graphics. $\endgroup$
    – lhf
    Oct 3, 2013 at 12:52

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