# Conditions on m for (M, p_m) to be a monoid

1. Sep 25, 2011

### farleyknight

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

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?