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abstract algebra - For $G$ group and $H$ subgroup of finite index ...
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How to prove $\\operatorname{Tr}(AB) = \\operatorname{Tr}(BA)$?
Jan 11, 2015 · The efficient @hjhjhj57: answer $$\text{Tr}(AB) = \text{Tr}(BA)= \sum a_{ij} b_{ji}$$ Now we can start to understand why if we do a circular permutation of the factors the expression