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"One can just work head-long into it: [a,b]=a−1b−1ab. $[a,b]=a^{-1}b^{-1}ab$. If there is no cancellation, you cannot have this equal to x−1y−1xyz−1w−1zw; $x^{-1}y^{-1}xyz^{-1}w^{-1}zw$; so either the last letter of a $a$ matches with the first of b, $b$, the first of a $a$ with the first of b $b$ (cancellation in b−1a), $b^{-1}a$), etc. Checking leads to impossibility. – Arturo Magidin" Can you give some hint on how to get contradiction? It seems to me too many cases to consider |
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"One can just work head-long into it: [a,b]=a−1b−1ab. If there is no cancellation, you cannot have this equal to x−1y−1xyz−1w−1zw; so either the last letter of a matches with the first of b, the first of a with the first of b (cancellation in b−1a), etc. Checking leads to impossibility. – Arturo Magidin" Can you give some hint on how to get contradiction? It seems to me too many cases to consider |
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