Abstract Algebra - Functions

Function

Classes of Functions

Injective

A function \(f:A \to B\) is called injective if each element of \(B\) is the image of no more than one element of \(A\). (单射)

Surjective

A function \(f: A \to B\) is called surjective if each element of \(B\) is the image of at least one element of \(A\). (满射)

Bijective

A function \(f: A \to B\) is called bijective if it is both injective and surjective, which defines a one to one mapping between \(A\) and \(B\).

Inverse

A function \(f:A \to B\) has an inverse if and only if it is a bijective.

Permutation

A bijective function \(f\) from \(A \to A\) is called a permutaion of a set \(A\).

Groups of permutation

All the permutation of \(A\), which the operation \(\circ\) of composition, is a group, where

1. The identity is \(\varepsilon\), where \(\varepsilon(x) = x, \forall \ x \in A\) .
2. The inverse is \(f^{-1}\) which satisfies \([f \circ f^{-1}](x) = \varepsilon(x)\).

Isomorphism

Let \(G_1\) and \(G_2\) be groups. A bijective function \(f: G_1 \to G_2\) with the property that for any two elements \(a\) and \(b\) in \(G_1\) that \(f(ab) = f(a)f(b)\) is called an isomorphism from \(G_1\) to \(G_2\).

💡 If there exists such an isomorphism \(f\), then we call \(G_1\) is isomorphic to \(G_2\), which is symbolized as \(G_1 \cong G_2\).

The logic behind the concise notation above is that \(f\) is bijective from \(G_1\) to \(G_2\). For \(a\) and \(b\) in \(G_1\), \(f\) maps \(a\) to \(a'\) in \(G_2\) and also \(b\). More importantly, in this context a same operation is discussed on \(G_1\) and \(G_2\), so \(f(ab)\) maps \(ab\) uniquely to \(a'b'\) in \(G_2\) since

\[ f(ab) = a'b' = f(a)f(b) \]

And another pre-requisite which needs to pay attention is that \(G_1\) and \(G_2\) are groups.

💡 Another attention have to be paid is that the operation on \(G_1\) and \(G_2\) can be different.

Cayley's Theorem

Every group is isomorphic to a group of permutations.

How to recognize isomorphism

1. Make an educated guess, and come up with a function \(f: G_1 \to G_2\) which looks as though it might be an isomorphism.
2. Check that \(f\) is bijective.
3. Check that \(f\) satisfies the identity \(f(ab) = f(a)f(b)\).

Automorphism

By an automorphism (自同构，自守) of \(G\) we mean an isomorphism \(f:G \to G\).

Partition And Equivalence Relations

Partition

By a partition of a set \(A\) we mean a family \(\{ A_i: i \in I \}\) of nonempty subsets of \(A\) such that

• If any two classes, say \(A_i\) and \(A_j\), have a common element \(x\) (that is, are not distinct), then \(A_i = A_j\).
• Every element \(x\) of \(A\) lies in one of the classes.

Equivalence Relation

By an equivalence relation on a set \(A\) we mean a relation \(\backsim\) which is

• Reflexive: \(x \backsim x\) for every \(x \in A\).
• Symmetric: if \(x \backsim y\), then \(y \backsim x\).
• Transitive: if \(x \backsim y\) and \(y \backsim z\), then \(x \backsim z\).

Equivalence Class

Let \(\backsim\) be an equivalence relation on \(A\) and \(x\) an element of \(A\). The set of all the elements equivalent to \(x\) is called the equivalence class of \(x\), and it's denoted by \([x]\). That is

\[ [x] = \{ y \in A: y \backsim x\} \]

Homomorphism

Homomorphism for Groups

If \(G\) and \(H\) are groups, a homomorphism from \(G\) to \(H\) is a function: \(f: G \to H\) such that for any two elements \(a\) and \(b\) in \(G\),

\[ f(ab) = f(a)f(b) \]

💡 If there exists a homomorphism from \(G\) onto \(H\), we say that \(H\) is a homomorphic image of \(G\).

Homomorphism for Rings

A homomorphism from a ring \(A\) to a ring \(B\) is a function \(f:A \to B\) satisfying the identities

\[ f(x_1 + x_2) = f(x_1)f(x_2) \\ f(x_1x_2) = f(x_1)f(x_2) \]

Kernel

Let \(f: G \to H\) be a homomorphism. The kernel of \(f\) is the set \(K\) of all the elements of \(G\) which are carried by \(f\) onto the neutral eleemnt of \(H\). That is,

\[ K = \{ x \in G: f(x) = e\} \]

💡 The kernel of a group \(G\) is a normal subgroup of \(G\). And the kernel of a ring \(A\) is an ideal of \(A\).

Basic Theorems

Let \(G\) and \(H\) be groups, and \(f: G \to H\) a homomorphism. Then

• \(f(e) = e\).
• \(f(a^{-1}) = [f(a)]^{-1}, \forall \ a \in G\).

Let \(f: G \to H\) be a homomorphism.

• The kernel of \(f\) is a normal subgroup of \(G\).
• The range of \(f\) is a subgroup of \(H\).

Let \(f: G \to H\) be a homomorphism with kernel \(K\). Then \(f(a) = f(b)\) iff \(Ka = Kb\).

Fundermental Homomorphism Theorem

FHT for Groups

Let \(f: G \to H\) be a homomorphism of \(G\) onto \(H\). If \(K\) is the kernel of \(f\), then

\[ H \cong G/K \]

💡 It asserts that every homomorphic image of \(G\) is isomorphic to a quotient group of \(G\). Which specific quotient group of \(G\) ? Well, if \(f\) is a homomorphism from \(G\) onto \(H\), then \(H\) is isomorphic to the quotient group of \(G\) by the kernel of \(f\).

FHT for Rings

Let \(f: A \to B\) be a homomorphism from a ring \(A\) onto a ring \(B\), and let \(K\) be the kernel of \(f\). Then

\[ B \cong A/K \]