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Modern Algebra: Basic problem dealing with Cosets
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[QUOTE="HallsofIvy, post: 4867754, member: 637751"] I wouldn't try to do it that way. I would, instead, use individual elements of H. And then use inverse of the elements of H, not a and b. "Ha= bH" means that, for any x in H, there exist y in H such that xa= by. Multiply on both sides, [B]on the right[/B], by [itex]y^{-1}[/itex] to get [itex]xay^{-1}= b[/itex]. Now multiply both sides, on the left, by [itex]x^{-1}[/itex]: [itex]ay^{-1}= x^{-1}b[/itex]. [/QUOTE]
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Modern Algebra: Basic problem dealing with Cosets
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