Category Theory: Monoids (Part I)
Point of Post: To give a perfunctory background into monoid theory, in preparation for their appearance as both an object of study, and method to study categories.
What is a monoid?
In essence a monoid is a group without inverses. Said differently, a monoid is just a set with an associative binary operation and an identity element. In fact, as we’ll see in the examples monoids come up quite often.
If you remember back to your days of group theory, you were often consternated by sets which just barely didn’t qualify as groups. More often than not, it was because they lacked inverses. with addition, with multiplication, with multiplication, etc. In all these cases (and much more) the structure is actually a monoid (a commutative monoid). Also, categories can be seen as somewhat of a generalization of monoids, in some sense. In fact, not only is every monoid a category (as was commented on in the last post) but there is a natural (not in the categorical sense) to turn a category into a monoid. We’ll discuss this later. So, let us give the definition
A monoid is a set , and a binary operation
a) is associative, in the sense that
for all .
b) There exists some element such that
Remark: It is common practice to drop the function notation in lieu of concatenation. Meaning, instead of having to write all the time, we simply let to mean . With this in mind the, admittedly menacing, axioms above become the friendly lookin
a) For all it’s true that
b) There exists some such that for all
So, we now state some examples so that the reader can see the diversity in which monoids appear
Along with the previous mentioned examples, and all groups, the following are monoids:
Example: Let be a set and let . We claim that where is normal function composition forms a monoid.
To see this we first must show that is really a well-defined map , but this is equivalent to showing that if are right invertible, then so is . But, this is clear since there exists some and some which map such that . So then, and
so that is indeed right invertible
Next, we need to check that is associative, but this is true since it’s just plain function composition. Thus, all that remains is to find an identity element, but since the identity map is right invertible we have that and evidently for all from where the conclusion follows.
Example: Let for any field . Then, where is the normal matrix multiplication is a monoid with identity element (the identity matrix)
Example: Let be a monoid, then with is a monoid.
Remark: This maybe isn’t so obvious since it’s clearly not true for groups. So, we’ll prove it.
We first note that clearly if then since we assumed that was closed under multiplication. So is indeed a binary operation.
Let and let then for some and . But, since we assumed that was a monoid it follows that and so . The reverse is done similarly: let , then for some and . But, since monoid multiplication is always associative we see that so that . It follows that so that the prescribed multiplication on is associative.
Furthermore, let be the identity element of , then and
The following theorem for monoids will prove useful, namely
Theorem: Let be a monoid, then the identity element of is unique
Proof: Let be another such element with the property that for all . Then, combining this with the same property for we get
from where the conclusion follows.