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Mar 23, 2012 at 9:39 vote accept domenico fiorenza
Mar 23, 2012 at 9:39 comment added domenico fiorenza Hi Gabriel, very nice answer, thanks! Let me summarize and simplify a bit your argument: Let M be a monoid; then: i) for each pair x,y of invertible elements there exist a homeomorphism of M in itself mapping x to y; ii) either M is a group or there exist an element which has neither a left inverse nor a right inverse; iii) if M is path connected but M-{e} is not connected then all elements of M except at most those in one connected component of M-{e} are invertible. Thus a monoid structure on the infinite ternary graph would give a homeomorphism mapping an edge internal point onto a vertex.
Mar 21, 2012 at 17:50 comment added Gabriel C. Drummond-Cole I played fast and loose with 2-sided invertibility, but it doesn't matter. Either: -there is a point which is not invertible. The argument goes through as above. -every element is invertible on one side or the other. Assume x is left invertible but not right invertible, and y is the reverse. xy is WLOG left invertible or right invertible so y is left invertible, a contradiction. Then WLOG every left invertible element is 2-sided invertible, so every right invertible element is, so every point in the monoid is 2-sided invertible.
Mar 21, 2012 at 10:43 history answered Gabriel C. Drummond-Cole CC BY-SA 3.0