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It is a basic fact that when two idempotents $e,f$ in a semigroup $S$ commute, then $ef$ is an idempotent. Is the converse true? Is it true for idempotents in rings?

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    @DavideGiraudo Yes.2012-03-04

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No. In $M_2(\mathbb{C})$, let $ e=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}, \ \ f=\begin{bmatrix}1&1\\ 0&0\end{bmatrix}. $ So $ e^2=e, \ \ f^2=f, \ \ ef=f, \ \ fe=e. $

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    I usually work with operators on a Hilbert space. In the case of projections, i.e. Selfadjoint idempotents, it is true that selfadjointness of a product of selfadjoints implies that they commute. So I just looked for the easiest non-selfadjoint idempotent, which is my $f$.2012-03-04
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Consider the semi-group of matrix $2\times 2$ with real entries and $A:=\pmatrix{0&1\\\ 0& 1}$, $B=\pmatrix{1&0\\\ 0&0}$. Then $A^2=A$, $B^2=B$, $AB=0$ but $BA=\pmatrix{0&1\\\ 0&0}\neq AB$.