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A monoid is $(M,*,e)$. I am having trouble understanding what the 'multiplication' operation represents. Is it an operation for an element which returns another element or an operation which takes two elements and returns a third one?

K.K
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  • You can think of it as regular old associative multiplication. For example the positive integers with multiplication. But it can represent other things, such as function composition or integer addition. – David P Feb 06 '14 at 21:31
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    Most "usual" monoids can be represented as functions with composition. But there are some monoids for which this is not true... – Thomas Andrews Feb 06 '14 at 21:40
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    You have $\star:M\times M\rightarrow M$. Denoting $\star\left(a,b\right)$ by $a\star b$ you have $a\star\left(b\star c\right)=\left(a\star b\right)\star c$. Element $e$ is characterized by $e\star a=a=a\star e$ for each $a\in M$. It can be shown that it is unique in this. – drhab Feb 06 '14 at 21:43

2 Answers2

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The second: it represents an associative binary operation which takes elements, say, $a, b\in M$ and operates on them, and outputs a third (not necessarily distinct from a or b) value:

$$a * b = c, \;\quad a, b, c\in M$$

The symbol $*$ here is a "generic" symbol representing a binary operation which needs to be defined or specified: for example, addition of integers $(a + b)$, multiplication of positive integers $(a\cdot b \;\text{or just}\; ab)$.

amWhy
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  • and $e$ would be something like a*e=a – K.K Feb 06 '14 at 21:33
  • Is it strictly on pairs of elements? Is it never the case that an element is treated individually? – K.K Feb 06 '14 at 21:35
  • Exactly. That's correct ($ae = a)$. Treated individually? NO: The "bi" in binary requires two* inputs. We might use, say, $a^2$ but that's just shorthand for $aa$. Note that we still need two elements here (which may not be distinct, as in $aa$) on which to perform the binary operation. – amWhy Feb 06 '14 at 21:35
  • Thank you. Could the set $M$ include the same elements more than once and maybe including the element $e$ more than once too as long as the rules are obeyed? Since the monoid forms a category too would the morphisms be morphisms in the category? – K.K Feb 06 '14 at 21:46
  • In a set, all elements are unique. That is, ${a, a, b, b, b, c, e} = {a, b, c, e}.$ But they can each be operated on repeatedly. – amWhy Feb 06 '14 at 21:48
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    I am typing some basics about these wild creatures in Abstract algebra. +1 – Mikasa Feb 07 '14 at 15:58
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The second is correct.

As an example you can consider the natural numbers $(\mathbb{N},+,0)$.

Anthony Conway
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