3. Identity Elements and Monoids, Inverses and Groups

Building up to Group Theory.

Let S be a semigroup and suppose there is an element 1 in S such that

\(\forall a\in S,\text{ }1\cdot a=a\cdot 1=a.\)

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

\(b=a^{-1}.\)

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

\(a^2=a.\)

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

\(\forall a\in G,\text{ }\exists a^{-1}\in G\text{ s.t. }aa^{-1}=a^{-1}a=1.\)

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,

\(\exists e^{-1}\in G\text{ s.t. } ee^{-1}=e^{-1}e=1.\)

Since 1 is the identity, we also have that

\(e^2=e=1\cdot e.\)

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

\(e=e^2\cdot e^{-1}=1\cdot e\cdot e^{-1}=1\)

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

\(\{a^n\in M: n\in\mathbb{Z}^{+}\}.\)

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

\(a^n=a^m.\)

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

\((a^k)^2=a^k.\)

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

\(\ell= (2k \mod m-n)-(k\mod m-n)\)

Then

\((k+\ell)\mod (m-n)=2(k+\ell)\mod (m-n)\)

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

First, observe that

\((a^6)^2=a^{12}=a^{10}\cdot a^2=a^4\cdot a^2=a^6\)

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

\((a^5)^2=a^{10}=a^4\)

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

\(a^k=1\Rightarrow a\cdot a^{k-1}=a^{k-1}\cdot a=1\)

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:

\(\forall a,b,c\in S,\text{ }ca=cb\text{ or }ac=bc\Rightarrow a=b. \)

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.

\(ca=cb\Rightarrow c^{-1}ca=c^{-1}cb\Rightarrow 1\cdot a=1\cdot b\Rightarrow a=b.\)

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

\(aS=\{as:s\in S\}.\)

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

\(a\neq b\Rightarrow ca\neq cb\)

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. □