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Q: If a and b are group elements and ab ≠ ba, prove that aba ≠ identity.

I began by stating my theorem, then assumed ab ≠ ba. Then I tried a few inverse law manipulations, which worked in a sense, however they brought me nowhere, as I couldn't conclude my proof concretely. Suggestions? Solutions?

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I recommend an approach by contrapositive. If $aba=e,$ then $a(ba)=e$ (so $ba=a^{-1}$) and $(ab)a=e$ (so $ab=a^{-1}$), and so....

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Try proving the contrapositive; that is, if $aba=e$ then $ab=ba$. First of all, multiply both sides of $aba=e$ on the left by $a^{-1}$ to get $ba=a^{-1}$; now multiply both sides of this on the right by $a^{-1}$ to get $b=a^{-2}$. Now, can you see why $b$ commutes with $a$?

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$$aba=1\implies ab=a^{-1}\;\;(1)\implies a=a^{-1}b^{-1}=(ba)^{-1}\;\;(2)$$

From (1)-(2) it follows

$$(1)\;a^{-1}=ab\implies a=(ab)^{-1}=b^{-1}a^{-1}\stackrel{(2)}=a^{-1}b^{-1}\ldots$$

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