2.3: Groups

Without further ado, here is our official definition of a group.

Definition: Group

A group \((G,*)\) is a set \(G\) together with a binary operation \(*\) such that the following axioms hold.

0. The set \(G\) is closed under \(*\).
1. The operation \(*\) is associative.
2. There is an element \(e\in G\) such that for all \(g\in G\), \(e*g=g*e=g\). We call \(e\) the identity*.

*

The origin of using the letter \(e\) for the identity of a group appears to be due to German mathematician Heinrich Weber, who uses “einheit" (German for “unit" or “unity") and \(e\) in his Lehrbuch der Algebra (1896).

3. Corresponding to each \(g\in G\), there is an element \(g'\in G\) such that \(g*g'=g'*g=e\). In this case, \(g'\) is said to be an inverse of \(g\).

The order of \(G\), denoted \(|G|\), is the cardinality of the set \(G\). If \(|G|\) is finite, then we say that \(G\) has finite order. Otherwise, we say that \(G\) has infinite order.

In the definition of a group, the binary operation \(*\) is not required to be commutative. If \(*\) is commutative, then we say that \(G\) is abelian*. A few additional comments are in order.

*

Commutative groups are called abelian in honor of the Norwegian mathematician Niels Abel (1802–1829).↩

• Axiom 2 forces \(G\) to be nonempty.
• If \((G,*)\) is a group, then we say that \(G\) is a group under \(*\).
• We refer to \(a*b\) as the product of \(a\) and \(b\) even if \(*\) is not actually multiplication.
• For simplicity, if \((G,*)\) is a group, we will often refer to \(G\) as being the group and suppress any mention of \(*\) whatsoever. In particular, we will often abbreviate \(a*b\) as \(ab\).
• In Theorem 2.3.3., we shall see that each \(g\in G\) has a unique inverse. From that point on, we will denote the inverse of \(g\) by \(g^{-1}\).

Problem \(\PageIndex{1}\)

Explain why Axiom 0 is unnecessary.

Problem \(\PageIndex{2}\)

Verify that each of the following is a group under composition of actions and determine the order. Which of the groups are abelian?

1. \(\text{Spin}_{3\times 3}\)
2. \(R_4\) (see Problem 2.2.4)
3. \(D_3\) (see Problem 2.2.5)
4. \(D_4\) (see Problem 2.2.6)
5. \(S_3\) (see Problem 2.2.7)

Problem \(\PageIndex{3}\)

Determine whether each of the following is a group. If the pair is a group, determine the order, identify the identity, describe the inverses, and determine whether the group is abelian. If the pair is not a group, explain why.

1. \((\mathbb{Z},+)\)
2. \((\mathbb{N},+)\)
3. \((\mathbb{Z},\cdot)\)
4. \((\mathbb{Z},\div)\)
5. \((\mathbb{R},+)\)
6. \((\mathbb{R},\cdot)\)
7. \((\mathbb{Q}\setminus \{0\},\cdot)\)
8. \((M_{2\times 2}(\mathbb{R}),+)\) Note: \(M_{2\times 2}(\mathbb{R})\) is the set of \(2\times 2\) matrices.
9. \((M_{2\times 2}(\mathbb{R}),*)\), where \(*\) is matrix multiplication.
10. \(([0,1],*)\), where \(a*b:=\min(a,b)\)
11. \((\{a,b,c\},*)\), where \(*\) is the operation determined by the table in Example 2.2.2.
12. \((\{x,y,z\},*)\), where \(*\) is the operation determined by the table in Problem 2.2.13.

Notice that in Axiom 2 of Definition: Group, we said the identity and not an identity. Implicitly, this implies that the identity is unique.

Theorem \(\PageIndex{1}\): Unique Identity

If \(G\) is a group, then there is a unique identity element in \(G\). That is, there is only one element \(e\in G\) such that \(ge=eg=g\) for all \(g\in G\).

Problem \(\PageIndex{4}\)

Provide an example of a group of order 1. Can you find more than one such group?

Any group of order 1 is called a trivial group. It follows immediately from the definition of a group that the element of a trivial group must be the identity.

It is important to note that if we have an equation involving the product of group elements, we can still “do the same thing to both sides" and maintain equality. However, because general groups are not necessarily abelian, we have to be careful that we truly operate in the same way on each side. For example, if we have the equation \(g = h\) in some group, then we also have \(ag = ah\), where we “multiplied" both sides on the left by the group element \(a\). We could not necessarily conclude that \(ag = ha\), unless one pair of the elements happen to commute with each other.

The following theorem is crucial for proving many theorems about groups.

Theorem \(\PageIndex{2}\): Cancellation Law

Let \(G\) be a group and let \(g,x,y\in G\). Then \(gx=gy\) if and only if \(x=y\). Similarly, \(xg=yg\) if and only if \(x=y\)*.

*

You only need to prove one of these statements as the proof of the other is similar.↩

Problem \(\PageIndex{5}\)

Show that \((\mathbb{R},\cdot)\) fails the Cancellation Law confirming the fact that it is not a group.

Recall that Axiom (3) of Definition: Group states that each element of a group has at least one inverse. The next theorem tells us that each element has exactly one inverse.

Theorem \(\PageIndex{3}\): Unique Inverse

If \(G\) is a group, then each \(g\in G\) has a unique inverse.

In light of the previous theorem, the unique inverse of \(g\in G\) will be denoted as \(g^{-1}\). In groups, it turns out that inverses are always “two-sided". That is, if \(G\) is a group and \(g,h\in G\) such that \(gh=e\), then it must be the case that \(hg=e\), as well. In this case, \(g^{-1}=h\) and \(h^{-1}=g\). However, there are mathematical structures where a “left inverse" exists but the “right inverse" does not.

Theorem \(\PageIndex{4}\): Unique Solution

If \(G\) is a group, then for all \(g,h\in G\), the equations \(gx=h\) and \(yg=h\) have unique solutions for \(x\) and \(y\) in \(G\).

The next theorem should not be surprising.

Theorem \(\PageIndex{5}\)

If \(G\) is a group, then \((g^{-1})^{-1}=g\) for all \(g\in G\).

The next theorem is analogous to the “socks and shoes theorem" for composition of functions.

Theorem \(\PageIndex{6}\)

If \(G\) is a group, then \((gh)^{-1}=h^{-1}g^{-1}\) for all \(g,h\in G\).

Definition: Exponents

If \(G\) is a group and \(g\in G\), then for all \(n\in \mathbb{N}\), we define:

1. \(g^n=\underbrace{gg\cdots g}_{n\text{ factors}}\)
2. \(g^{-n}=\underbrace{g^{-1}g^{-1}\cdots g^{-1}}_{n\text{ factors}}\)
3. \(g^0=e\)

Remark \(\PageIndex{1}\)

If \(G\) is a group under \(+\), then we can reinterpret Definition: Exponents as:

1. \(ng=\underbrace{g+g+\cdots +g}_{n\text{ summands}}\)
2. \(-ng=\underbrace{-g+-g+\cdots +-g}_{n\text{ summands}}\)
3. \(0g=0\)

Notice all that we have done is taken the statements of Definition: Exponents, which use multiplicative notation for the group operation, and translated what they say in the case that the group operation uses additive notation.

The good news is that the many of the rules of exponents you are familiar with still hold for groups.

Theorem \(\PageIndex{7}\): Exponents

If \(G\) is a group and \(g\in G\), then for all \(n,m\in\mathbb{Z}\), we have the following:

1. \(g^ng^m=g^{n+m}\),
2. \((g^n)^{-1}=g^{-n}\),
3. \((g^n)^{m}=g^{nm}\).

Problem \(\PageIndex{6}\)

Reinterpret Theorem \(\PageIndex{7}\) if \(G\) is a group under addition.

Unfortunately, there are some rules of exponents that do not apply for general groups.

Problem \(\PageIndex{7}\)

Show with a specific example that for a group \(G\) we may have \((ab)^2\neq a^2b^2\). What property would guarantee that \((ab)^2=a^2b^2\) for all \(a,b\in G\)? Is the converse of your claim true?