Introduction to Abstract Algebra

投稿日:2018/5/16最終更新日:2025/5/17

A brief introduction to Abstract Algebra

Introduction to Groups

Definitions and Examples

Algebra is essentially the study of sets equipped with various binary operations.

Definition 1

Let SS be a set. A binary operation on SS is defined as a mapping :S×SS* : S \times S \rightarrow S. Furthermore, for a,bSa, b \in S, we write (a,b)=ab*(a, b) = a * b. The above definition explains that an operation * on a set SS is called a binary operation if for every a,bSa, b \in S, the result aba * b is also an element of SS. Some familiar examples of binary operations are +,×,+, \times, - on the set of real numbers R\mathbb{R} (or on the set of rational numbers Q\mathbb{Q}, or the set of integers Z\mathbb{Z}). Likewise, division on the set R{0}\mathbb{R} \setminus \{0\} or Q{0}\mathbb{Q} \setminus \{0\} is a binary operation. However, division on the set Z{0}\mathbb{Z} \setminus \{0\} is not a binary operation because the result of the operation is not always an element of Z{0}\mathbb{Z} \setminus \{0\}. For example, 1,2Z{0}1, 2 \in \mathbb{Z} \setminus \{0\}, but their quotient 12Z{0}\frac{1}{2} \notin \mathbb{Z} \setminus \{0\}.

In the case of groups, we focus on studying sets that are equipped with a single binary operation that satisfies certain specific characteristics. A common initial example of a group comes from the sets Z,Q,R\mathbb{Z}, \mathbb{Q}, \mathbb{R}, and the set of complex numbers C\mathbb{C}. In these sets, the operations of addition (++) and multiplication (×\times) are binary operations. Subtraction and division can also be described simply in terms of addition and multiplication. For example, 57=5+(7)5 - 7 = 5 + (-7) and
5÷7=5×175 \div 7 = 5 \times \frac{1}{7}.

The operations of addition and multiplication, which are binary operations on these number sets, possess several interesting characteristics. For instance, the addition operation on Z\mathbb{Z} satisfies the following axioms:

  1. There exists an integer 00, such that for every integer xx, x+0=0+x=xx + 0 = 0 + x = x. We call the integer 00 the additive identity.

  2. For every integer xx, x-x is an integer that is the additive inverse of xx, since x+(x)=(x)+x=0x + (-x) = (-x) + x = 0.

  3. For every three integers x,y,zx, y, z, x+(y+z)=(x+y)+zx + (y + z) = (x + y) + z. This property is called the associativity of addition.

It can be checked that the axioms of the addition operation also apply to the sets Q,R,\mathbb{Q}, \mathbb{R}, and C\mathbb{C}.

However, for the multiplication operation, we need to be a bit more careful because multiplication on Z\mathbb{Z} does not satisfy the third point — for example, the element 2Z2 \in \mathbb{Z} does not have a multiplicative inverse in Z\mathbb{Z}, since 12Z\frac{1}{2} \notin \mathbb{Z}. What about the sets Q,R,\mathbb{Q}, \mathbb{R}, and C\mathbb{C}? Do multiplication operations on these sets satisfy all four of the previous points? Once again, the third point fails because the number 00 does not have an inverse. Therefore, by restricting the sets Q,R,\mathbb{Q}, \mathbb{R}, and C\mathbb{C} to only their nonzero elements, denoted respectively by Q,R,\mathbb{Q}^*, \mathbb{R}^*, and C\mathbb{C}^*, the multiplication operation on these sets satisfies all the axioms above. For example, consider the set Q\mathbb{Q}^* with the multiplication operation ×\times:

  1. There exists a rational number 11, such that for every xQx \in \mathbb{Q}^*, x×1=1×x=xx \times 1 = 1 \times x = x. The rational number 1 is called the multiplicative identity.

  2. For every xQx \in \mathbb{Q}^*, x1=1xQx^{-1} = \frac{1}{x} \in \mathbb{Q}^* is the multiplicative inverse of xx, because it satisfies x×1x=1x×x=1x \times \frac{1}{x} = \frac{1}{x} \times x = 1.

  3. For all x,y,zQx, y, z \in \mathbb{Q}^*, the associative property holds: x×(y×z)=(x×y)×zx \times (y \times z) = (x \times y) \times z.

From the previous examples, we can now observe the formal definition of a group as an abstraction from a pair (G,)(G, \ast), where GG is a set and :G×GG\ast : G \times G \rightarrow G is a binary operation satisfying the following axioms.

Definition 2

A non-empty set GG with a binary operation :G×GG* : G \times G \rightarrow G is called a group if:

  1. The associative property holds: a(bc)=(ab)ca * (b * c) = (a * b) * c for all a,b,cGa, b, c \in G.

  2. There exists an identity element eGe \in G, such that for every aGa \in G, ea=ae=ae * a = a * e = a.

  3. Every element aGa \in G has an inverse: there exists bGb \in G such that ab=ba=ea * b = b * a = e. This element bb is called the inverse of aa, and is denoted by a1a^{-1}.

If (G,)(G, *) is a group and ab=baa * b = b * a for all a,bGa, b \in G, then (G,)(G, *) is called a commutative group or an abelian group.

In the definition above, the associative property of * provides freedom in the order of performing operations on several elements. In some other sources, a condition is also mentioned that the operation * must be closed over the set GG (i.e., for all a,bGa, b \in G, abGa * b \in G). However, this closure condition is already implicitly satisfied by the assumption that * is a binary operation as defined in Definition 1.

Some examples of groups.

In the previous discussion, we have already seen some common examples of groups. In addition to those, some other examples can be seen below.

Example 3. The set of positive rational numbers Q+\mathbb{Q}^+ and the set of positive real numbers R+\mathbb{R}^+ form groups under the operation of multiplication ×\times, with identity element e=1e = 1 and inverse a1=1aa^{-1} = \frac{1}{a} for each positive rational or real number. Clearly, if aQ+a \in \mathbb{Q}^+, then a1Q+a^{-1} \in \mathbb{Q}^+. The same holds for R+\mathbb{R}^+.

Addition operations do not always form a binary operation, depending on the set used. Consider the following example.

Example 4. The set of all odd numbers G={2k+1kZ}G = \{2k + 1 \mid k \in \mathbb{Z}\} under addition does not form a group. This is because addition is not a binary operation (or not closed) on GG:

(2k+1)+(2l+1)=2(k+l)+2G.(2k + 1) + (2l + 1) = 2(k + l) + 2 \notin G.

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