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Conditions on m for (M, p_m) to be a monoid
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[QUOTE="farleyknight, post: 3522030, member: 159443"] [h2]Homework Statement [/h2] Let (M, *) be a monoid. Define p_m(a, b) = amb. The first part of this question was to prove that (M, p_m) defines a semigroup. The second part is to suggest what conditions on m are needed for (M, p_m) to be a monoid. [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] If we would like (M, p_m) to be a monoid, we would require that the element m in M satisfy: a = p_m(a, 1_M) = am * 1_M = am = ma = 1_M * p_m(1_M, a) In other words, a = ma = am must hold for all a in M. However, this is exactly the condition for m to be the unit of the monoid (M, p_m). So (M, p_m) is a monoid iff m = 1_M My previous attempt only included am = ma, meaning that m would have to only commute with a, which is obviously incorrect. My teacher is allowing for a re-write of this question and suggests that m must only be invertible. However, that doesn't seem to be enough.. Or am I missing something? [/QUOTE]
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Conditions on m for (M, p_m) to be a monoid
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