Then we say that 1 is the identity element of S. I write 1 for notational purposes only (meaning 1 isn’t necessarily the number 1—just a symbol representing the identity element). If S contains an identity element, then we say that S is a monoid.

Fact: The identity element of a monoid is unique.

Proof: Exercise.

Let a be in a monoid M with identity 1. We say that an element b is an inverse of a if ab=ba=1 and we write

Fact: The inverse of an element in a monoid is unique.

Proof: Exercise.

An idempotent is an element a in a semigroup (or monoid) such that

The identity element of a monoid is an idempotent. I now give the fundamental definition for this series.

Definition: A group is a monoid such that every element has an inverse.

That is, a group is a monoid G such that

Theorem: The only idempotent in a group G is the identity.

Proof: Suppose to the contrary that e is an idempotent in G. Since G is a group e has an inverse. That is,

Since 1 is the identity, we also have that

Then by left multiplication of the inverse of e on both sides of the =, we obtain

so e=1. □

Theorem: Let M be a finite monoid with a unique idempotent. Then M is a group.

Proof: Since the identity 1 is always idempotent, we have that the unique idempotent must be the identity. Consider the set P of all powers of some a in M

P is a subsemigroup of M since it is closed under multiplication. (in fact it is a subsemigroup of M (the “monogenic“ subsemigroup of M generated by a) since it is closed under multiplication. Hence P must be finite, so there must exist some positive integers n and m (with n less than or equal to m) such that

I claim that there exists some k which is greater than or equal to n and less than or equal to m such that

That is, a to the power of k is idempotent. Why? Suppose to the contrary that there is no such k. Let k be between n and m as mentioned before. By assumption, 2k mod m-n is not equal to k mod m-n. Let

Then

so a to the power of k+l is idempotent. The below figure illustrates this argument more clearly.

First, observe that

so a to the power of 6 is an idempotent. Now choose an arbitrary k, lets say k=5. From the diagram, we see that

so we have fallen “1 a short“ of reaching around the loop to get back to a to the power of five by just multiplying by 5. We then would try a to the power of 6 next, since 6 is one greater than 5. This must work because we now walk 2 a’s further around the circle when squaring a to the power of 6, but we also must stretch an additional one a around to get to a to the power of 6. Try again with k=7 and we see that when walking a distance of 7 around the circle, we go one too far, so we should try k=7-1=6, which we know works.

Now that that is settled, return to the question at hand: showing that every element has an inverse. Luckily, now that this fact has been established, this is easy. For every a in M, there is some power k of a which is idempotent. We then get the following implication

so the inverse of a is a to the power of k-1. Then every element a in M has an inverse, so M is a group! □

I now give an important characterization of groups.

Theorem: A semigroup G is a group if and only if for any a and b in G, there exist x and y in G such that ax=b and ya=b.

Proof: Exercise.

We say that a semigroup S is cancellative if it has the following property:

That is, you can strike the same element off of both sides of an equals sign and preserve equality. All groups are cancellative, since multiplication on both sides of an equals sign by the inverse of c and reducing will result in a=b.

We then have the following theorem.

Theorem: A finite cancellative semigroup S is a group.

Proof: Let a be in S. I use the following notation

Sa is defined similarly with multiplication on the right. By the contrapositive of the cancellation property, we have that

so |aS| is greater than or equal to |S|. Since S is closed, we have that |aS|=|S|. Similarly, |Sa|=|S|. Then for any b in S, there exists some x and y in S such that ax=b and ya=b. □

is called a binary operation on A and a subset B of A is said to be closed under the binary operation if for any a and b in B ∗(a,b) is in B. We usually write a∗b or even ab (and we say “a times b“ for a generic operation) for ∗(a,b). A magma is a set A equipped with a binary operation ∗, written (A,∗) or simply A when ∗ is understood.

An operation on A is said to be associative if for any a, b, and c in A,

In this case, we call A a semigroup. if B is a subset of A which is closed under the same operation as A, then we call B a subsemigroup of A.

It is important to note that associativity in no way implies commutativity, the property where for any a and b in A, ab=ba. I now give some examples of semigroups. It is left to the reader to prove that they are indeed semigroups.

• The set of real numbers under addition or multiplication

• The set of integers under addition/multiplication is a subsemigroup of the real numbers under addition/multiplication

• The set of even integers under addition or multiplication is a subsemigroup of the set of integers under addition/multiplication

• The set of all functions from a set to itself

• The set of all bijections from a set to itself is a subsemigroup of the set of all functions from a set to itself

• Square matrices under matrix multiplication or addition

Notably, the set of odd integers is NOT a semigroup, since 1+1=2, which is not odd. Finite semigroups may be described with a multiplication table where each entry represents the product of its row label (on the left) by its column label (on the right). A multiplication table can be checked for associativity via Light’s test.

Exercise: Consider the set S composed of rock, paper, and scissors defined by the below multiplication table. Is S a semigroup? Is S commutative?

We can also use the Cartesian product to construct a new semigroup from two existing semigroups.

Proposition: Let S and T be semigroups (written multiplicatively). Then the cartesian product of S and T is a semigroup with the following multiplication (s’s are in S and t’s in T):

Proof: It is evident from the definition that the Cartesian product is closed under multiplication and that multiplication is defined for all pairs. It then remains to check that multiplication is associative. We compute:

and

Since multiplication in S and T is associative, we have that

so multiplication in the cartesian product is associative. □

Sets

A set is a well-defined collection of objects that can be described precisely, perhaps by a shared property. For two sets A and B, we say that A is a subset of B and write

if

If A is a subset of B, but

then we write

and say that A is a proper subset of B.

Cartesian Product

For two sets A and B, we denote the Cartesian product of A and B to be

For a finite collection of sets (such that i is bounded by 1 and n), we may write

For an arbitrary collection of sets indexed by a set I, we may write

Relations

A relation (between sets A and B) is a subset of the Cartesian product (of sets A and B). For a relation

if

then we may for convenience write

and say that a and b are related under R. One of the most commonly encountered types of relations is that of a function. A function f with domain A and codomain B, written

is a subset

That is, any a in A cannot be related to two distinct elements in B. In secondary school algebra, one learns that a function (from the Real line to itself) must “pass the vertical line test“ to be a function. This just simply means that any real number cannot be sent to two distinct real numbers under a function. We also use the following notation, which may also be used for relations in general.

Reframed in this notation, the definition of a function is familiar.

The range of f is the set of all b’s in B that are mapped onto by the function.

Just like functions, all relations can be composed. Formally, the composition of relations R and S is defined below. Let R be a relation from A to B and S be a relation from B to C.

All this means is that whenever R(a)=b and S(b)=c then

You may notice that function and relation composition is written “backwards“ relative to how we think about it. This is done to fit with the f(a)=b notation, however some authors write composition the “correct way“.

A function f from A to B is called one-to-one or injective if

It is called onto or surjective if

A function which is both one-to-one and onto is called a one-to-one correspondence or a bijection.

We denote the cardinality of a set A as |A| and we define

if there exists an injection from A to B or if there exists a surjection from B to A. Hence two sets have equal cardinality if there exists a bijection between them. If a set has finitely many elements n, then it is in bijection with the set of positive integers less than or equal to n, and we may say that it has cardinality n.

The identity map on a set A is the function

and is denoted id (when A is understood). Let

and

be functions. g is called a left (resp. right) inverse of f if

Theorem: A function f has a left inverse iff it is injective and a right inverse iff it is onto.

Proof: Exercise.

Equivalence Relations

Consider the following relation on a set A.

We say that ~ is reflexive if

symmetric if

and transitive if

A curious reader is encouraged to consider a finite set A and find a way to represent reflexive and symmetric relations in terms of a binary matrix. How could one interpret matrix multiplication in this context?

A partition of a set A is a collection of disjoint subsets of A such that their union equals A.

Theorem: Every equivalence relation defines a partition and every partition defines an equivalence relation.

Otherwise stated, every equivalence relation is a partition and every partition is an equivalence relation.

Proof: (⟹) Let R be an equivalence relation and for all a’s in A, let the equivalence class of a be the set of all elements which are related to a and denote it [a]. From reflexivity, every a is contained in some equivalence class. Hence the union of all such classes contains A. We also see that by symmetry, if b is in [a], then a is in [b], so [a]=[b] must be true. Suppose that for two elements a and b in A, [a]≠[b] and suppose to the contrary that there exists some c in the intersection of [a] and [b]. By transitivity, a and b must be related, so [a]=[b], which is a contradiction. Then each equivalence class is disjoint.

(⟸) Consider a partition of A into disjoint subsets. We construct a relation R such that for each such subset P and each a and b in P, R contains (a,b) and (b,a). Since every a in A is contained in some subset in the partition, (a,a) is in R, so R is reflexive. If (a,b) is in R, then a and b must be in the same subset P, so naturally a is in the same subset as b and therefore (b,a) is in R. Then R is symmetric. Now suppose that both (a,b) and (b,c) are in R. Then a ,b, and c must all be in the same subset, so (a,c) is in R. Then R is transitive and we may conclude that it is an equivalence relation. □

That sums up the preliminaries section. We may now move on to Algebra….

Below is an outline of topics:

1. preliminaries: a reminder of naïve set theory, and definitions of relations, functions, and the cartesian product

2. operations, closure, and semigroups

3. identity elements and monoids, inverses and groups

4. generators and Cayley graphs

5. Green’s relations, and the ax=b & ya=b definition of a group

6. monogenic semigroups and cyclic groups

7. homomorphisms, congruences, and isomorphisms

8. the First Isomorphism Theorem

9. transformation semigroups and symmetric groups

10. actions on a set

11. important theorems: Cayley, Lagrange, Cauchy

12. free objects and presentations

The outline may be changed during development to avoid circular logic and for the convenience of the reader, as I am not directly mimicking any particular textbook.