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Elementary property of cosets
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[QUOTE="fresh_42, post: 6036516, member: 572553"] This is hard to correct, because it is so easy, that obvious and for reason is hard to distinguish. Why? From ##ab^{-1}\in H ## we get ##ab^{-1}=h## for some ##h\in H##. Then ##a=ab^{-1}b=hb\in Hb##. But why ##abb^{-1}\,.## And Abelian isn't an argument, for a) you haven't required it and b) it isn't required at all. It looks as if you used what you wanted to show. I guess ##e_h=e_H\,.##Yes, and therefore ##ab^{-1}=(xb)b^{-1}=x\in H.## I don't understand the rest. [/QUOTE]
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Elementary property of cosets
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