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We define $_BA_B=(A,+_A, {\times}_L,{\times}_R)$ such that $(A,+_A,{\times}_L)$ is a left $B$-module and $(A,+_A,{\times}_R)$ is a right $B$-module. Provided that $b_1{\times}_L(a{\times}_Rb_2)=(b_1{\times}_La){\times}_Rb_2$ this is a bimodule.

Can there exist a situation where $b_1{\times}_La{\neq}a{\times}_Rb_1$, or would this not be a bimodule anymore if this equation was true? If the latter, how does it follow from bimodule's definition that it's commutative?

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Just take $A$ as your favourite non-commutative ring, for example the ring of quaternions, treated as a left and right module over itself. Then $\times_L$ and $\times_R$ are both just multiplication in $A$, and that $b_1{\times}_L(a{\times}_Rb_2)=(b_1{\times}_La){\times}_Rb_2$ follows from associativity of $A$. But $b_1{\times}_La{\neq}a{\times}_Rb_1$ because $A$ is not commutative, so we can find some pair of elements $a,b$ such that $ab \ne ba$.

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