Could someone help me with this proof? Thanks! Let G be a group, and H a subgroup of G. Ha = Hb iff ab^-1 is in H
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Originally Posted by chris27 Could someone help me with this proof? Thanks! Let G be a group, and H a subgroup of G. Ha = Hb iff ab^-1 is in H $\displaystyle Ha = Hb \Leftrightarrow Hab^{-1} = Hbb^{-1} \Leftrightarrow Hab^{-1} = He \Leftrightarrow Hab^{-1} = H \Leftrightarrow ab^{-1} \in H$.
I do not like your proof Soltras. $\displaystyle Ha = Hb$ now $\displaystyle a\in Ha$ so $\displaystyle a\in Hb$ that means $\displaystyle a=hb$ for some $\displaystyle ab^{-1} = h \in H$. The converse if almost identity.
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