Groups, Rings, and Fields

Instructor

Li Hui

Semigroups and Monoids

Semigroups

Definition: Let $S$ be a nonempty set, and let $*$ be a binary algebraic operation defined on $S$. If $*$ satisfies the associative law, then the algebraic system $U={\langle}S, *{\rangle}$ is called a semigroup.

When $S$ is a finite set, $U={\langle}S, *{\rangle}$ is called a finite semigroup.

When $S$ is an infinite set, $U={\langle}S, *{\rangle}$ is called an infinite semigroup.

For example, ${\langle}N, +{\rangle}$, ${\langle}N, \times{\rangle}$, ${\langle}E_v, +{\rangle}$, ${\langle}E_v, \times{\rangle}$, etc., are all semigroups.

Example: Define two binary operations $*$ and $\div$ on $R^+$:

$a*b=a^b$, and $\div$ denotes ordinary division.

This gives two algebraic systems $U={\langle}R^+, *{\rangle}$ and $V={\langle}R^+, \div{\rangle}$. Determine whether they are semigroups.

Solution: For $2, 3, 4{\in}R^+$, we have \[(2*3)*4 = 2^3*4 = (2^3)^4= 2^{12}\] \[2*(3*4) = 2*(3^4)=2*(81) =2^{81}\] Thus $(2*3)*4{\neq}2*(3*4)$.

By the properties of $\div$, it does not satisfy the associative law.

Therefore, neither $*$ nor $\div$ satisfies the associative law, so neither $U$ nor $V$ is a semigroup.

Example: Let $S_k=\{x|x{\in}Z, \text{ and } x{\geq}k\}$, $k{\geq}0$. Can $S_k$ with $+$ form a semigroup? Here $+$ denotes ordinary addition.

Solution: Closure. For any $x, y{\in}S_k$: $x{\in}Z$, $y{\in}Z$, $x{\geq}k$, $y{\geq}k$. So $x+y{\in}Z$, $x+y{\geq}k$.

Therefore $+$ is closed on $S_k$, and ${\langle}S_k, +{\rangle}$ is an algebraic system.

Associativity. Since $x+(y+z)=(x+y)+z$, $+$ satisfies the associative law.

Therefore ${\langle}S_k, +{\rangle}$ is a semigroup.

For this example, if the condition $k{\geq}0$ is removed, is ${\langle}S_k, +{\rangle}$ still a semigroup?

Example: Let $S_k=\{x|x{\in}Z, \text{ and } x{\geq}k\}$. Can $S_k$ with $+$ form a semigroup? Here $+$ denotes ordinary addition.

If $k<0$, take $k=-5$ without loss of generality.

Take two elements in $S_k$: $-2, -4{\in}Z$, clearly $-2, -4{\geq}k$, but $-2+(-4)=-6<k$, so $-6{\notin}S_k$.

Therefore $+$ is not closed on $S_k$, ${\langle}S_k, +{\rangle}$ is not an algebraic system, and ${\langle}S_k, +{\rangle}$ is not a semigroup.

Theorem: Let $U={\langle}S, *{\rangle}$ be a semigroup and let $X{\subseteq}S$ be a nonempty set. If $*$ is closed on $X$, then $V={\langle}X, *{\rangle}$ is also a semigroup.

This theorem shows that when closure holds, the associative law is preserved under restriction to a subset.

Definition: If the operation $*$ in a semigroup $U={\langle}S, *{\rangle}$ satisfies the commutative law, then $U$ is called a commutative semigroup.

For example, ${\langle}N, +{\rangle}$ and ${\langle}N, {\times}{\rangle}$ are both commutative semigroups, since both $+$ and ${\times}$ satisfy the commutative law.

Monoids

Definition: A semigroup containing an identity element $e$ is called a monoid.

For example, $U={\langle}N, +{\rangle}$ and $V={\langle}N, {\times}{\rangle}$ are both monoids, with 0 and 1 being the identity elements of $U$ and $V$, respectively.

Although ${\langle}N^*, +{\rangle}$ is a semigroup, there is no identity element for $+$, so ${\langle}N^*, +{\rangle}$ is not a monoid.

${\langle}E_v, {\times}{\rangle}$ is also not a monoid, since ${\times}$ has no identity element in $E_v$.

To emphasize the identity element $e$, a monoid ${\langle}S, *{\rangle}$ is often written as ${\langle}S, *, e{\rangle}$.

Definition: If $*$ in a monoid $U={\langle}S, *{\rangle}$ satisfies the commutative law, then $U$ is called a commutative monoid.

For example, ${\langle}Z, +{\rangle}$ is a commutative monoid.

As another example, for the power set $P(S)$ of any set $S$, with set operations ${\cap}$ and ${\cup}$, $U={\langle}P(S), {\cap}{\rangle}$ and $V={\langle}P(S), {\cup}{\rangle}$ are both commutative monoids, where $S$ is the identity of $U$ and $\emptyset$ is the identity of $V$.

Example: Let $R$ be the set of real numbers, define the operation $*$ ${\forall}a, b{\in}R$, $a*b=a+b+ab$.

Prove: $*$ forms a monoid on $R$.

Proof: Closure. For any $a, b{\in}R$: $a*b=a+b+ab$, since $a+b+ab{\in}R$, thus $a*b{\in}R$.

Associativity. For any $a, b, c{\in}R$: \[(a*b)*c=(a+b+ab)*c\] \[=(a+b+ab)+c+(a+b+ab)c\] \[=a+b+c+ab+ac+bc+abc\]

\[a*(b*c)=a*(b+c+bc)\] \[=a+(b+c+bc)+a(b+c+bc)\] \[=a+b+c+ab+ac+bc+abc\] Therefore $(a*b)*c=a*(b*c)$.

Identity element. Let ${\exists}e_l, e_r{\in}R$, for ${\forall}a{\in}R$: \[e_l*a=e_l+a+e_la=a\] We get $e_l(a+1)=0$, so $e_l=0$.

\[a*e_r=a+e_r+ae_r=a\] We get $e_r(a+1)=0$, so $e_r=0$.

Therefore $e_l=e_r=e=0$.

Commutativity. For any $a, b{\in}R$: \[a*b=a+b+ab=b+a+ba=b*a\] Therefore ${\langle}R, *{\rangle}$ is a commutative monoid.

Definition: Let $U={\langle}S, *{\rangle}$ be a monoid. If there exists an element $g{\in}S$ such that for all $a{\in}S$, $a$ can be expressed as a power of $g$, i.e., $a=g^n$ ($n$ is a natural number, with $e=g^0$ by convention), then $U$ is called a cyclic monoid.

$ is called a generator of $U$, and $U$ can be denoted $U={\langle}{\langle}g{\rangle}, *{\rangle}$.

For example, the set of natural numbers $N$ with ordinary addition $+$: ${\langle}N, +{\rangle}$ is a cyclic monoid, with identity $e=0$ and generator 1.

$1^0=0$, $1^1=1$, $1^2=1+1=2$, ${\cdots}$, $1^n=1+1+{\cdots}+1+1=n$, ${\cdots}$

We can write: ${\langle}N, +{\rangle}={\langle}{\langle}1{\rangle}, +{\rangle}$.

Example: Let $X=\{0, 60, 120, 180, 240, 300\}$, ${\oplus}$ be a binary operation on $X$, for ${\forall}a, b{\in}X$: \[a{\oplus}b=(a+b)\, \mod\, 360\] Then the algebraic system ${\langle}X, {\oplus}{\rangle}$ is a cyclic monoid.

Proof: Since ${\langle}X, {\oplus}{\rangle}$ is an algebraic system, closure holds.

Moreover, modular arithmetic satisfies the associative law, i.e., for ${\forall}a, b, c{\in}X$: $(a{\oplus}b){\oplus}c=a{\oplus}(b{\oplus}c)$, Therefore ${\langle}X, {\oplus}{\rangle}$ is a semigroup.

This semigroup has identity element 0, so ${\langle}X, {\oplus}{\rangle}$ is a monoid.

There is a generator $g=60$, because: \[g^0=0, g^1=60, g^2=120, g^3=180, g^4=240, g^5=300, g^6=0,\] \[g^7=60, g^8=120, g^9=180, g^{10}=240, g^{11}=300, {\cdots},\] Therefore ${\langle}X, {\oplus}{\rangle}$ is a cyclic monoid.

Example: Let monoid $U={\langle}S, *{\rangle}$, $S=\{1, a, b, c, d\}$, with $*$ defined as follows:

From this we can see that 1 is the identity element, $c$ is a generator, and: \[c^0=e=1, c^1=c, c^2=b, c^3=a, c^4=d\] \[ \begin{array}{cccccc} * & 1 & a & b & c & d\\ 1 & 1 & a & b & c & d\\ a & a & a & b & d & d\\ b & b & b & d & a & a\\ c & c & d & a & b & b\\ d & d & d & a & b & b\\ \end{array} \]

Theorem: Every cyclic monoid is a commutative monoid.

Proof: Let $U={\langle}S, *{\rangle}$ be a cyclic monoid with generator $G$.

For any $a, b{\in}S$, by the definition of the generator, there exist $m, n{\in}N$ such that: $a=g^m, b=g^n$

Then: \[a*b=g^m*g^n=g^{m+n}=g^{n+m}=g^n*g^m=b*a\] Therefore, $U$ is a commutative monoid.

Subsemigroups and Submonoids

Definition: Let $U={\langle}S, *{\rangle}$ be a semigroup and let $X{\subseteq}S$ be a nonempty subset. If $V={\langle}X, *{\rangle}$ also forms a semigroup, then $V$ is called a subsemigroup of $U$.

For example, ${\langle}E_V, {\times}{\rangle}$ is a subsemigroup of ${\langle}Z, {\times}{\rangle}$, and ${\langle}Z, +{\rangle}$ is a subsemigroup of ${\langle}R, +{\rangle}$.

Given a semigroup $U={\langle}S, *{\rangle}$ and a nonempty subset $X{\subseteq}S$, to verify that $V={\langle}X, *{\rangle}$ is a subsemigroup of $U$, we need $V$ to be a semigroup, i.e., $*$ must be a binary operation on $X$ (satisfying closure) and satisfy the associative law.

Since $*$ satisfies the associative law on $S$, as long as $*$ is closed on $X$, the associative law will hold on $X$ automatically.

Therefore, to prove that $V={\langle}X, *{\rangle}$ is a subsemigroup of $U={\langle}S, *{\rangle}$, it suffices to show:

$X{\subseteq}S$ is nonempty.
$*$ is closed on $X$.

Example: $U={\langle}Z, +{\rangle}$ is a semigroup, let: $Z_E=\{x|x=2n, n{\in}Z\}$,

Prove: ${\langle}Z_E, +{\rangle}$ is a subsemigroup of ${\langle}Z, +{\rangle}$.

Proof: Clearly $Z_E$ is a nonempty subset of $Z$.

Let ${\forall}x, y{\in}Z_E$, write $x=2n$, $y=2m$ with $n, m{\in}Z$, then $n+m{\in}Z$, and $x+y=2n+2m=2(n+m){\in}Z_E$, so closure holds, thus ${\langle}Z_E, +{\rangle}$ is a semigroup.

Therefore ${\langle}Z_E, +{\rangle}$ is a subsemigroup of ${\langle}Z, +{\rangle}$.

Definition: Let $U={\langle}S, *, e{\rangle}$ be a monoid and let $X{\subseteq}S$ be a nonempty subset. If $V={\langle}X, *, e{\rangle}$ also forms a monoid, then $V$ is called a submonoid of $U$.

It is particularly important that the identity element of $V$ must coincide with that of $U$. Otherwise, even if $V$ forms a monoid, it is not a submonoid of $U$.

For example, for the set of real numbers $R$, integers $Z$, natural numbers $N$ with ordinary addition $+$: ${\langle}R, +{\rangle}$ is a monoid with identity 0, and both ${\langle}Z, +{\rangle}$ and ${\langle}N, +{\rangle}$ are submonoids of ${\langle}R, +{\rangle}$.

Example: Let ${\langle}Z, +{\rangle}$ be a monoid, let: \[Z_E=\{x|x=2n, \, n{\in}Z\}, \] Prove: ${\langle}Z_E, +{\rangle}$ is a submonoid of ${\langle}Z, +{\rangle}$.

Proof: We have already proved that ${\langle}Z_E, +{\rangle}$ is a subsemigroup of ${\langle}Z, +{\rangle}$.

Now we show that ${\langle}Z_E, +{\rangle}$ has an identity element, which coincides with that of ${\langle}Z, +{\rangle}$. The identity element of ${\langle}Z, +{\rangle}$ is 0.

Clearly $0{\in}Z_E$. For any $x{\in}Z_E$, write $x=2n$:

$x+0=2n+0=2n=x$,

$0+x=0+2n=2n=x$,

Therefore 0 is the identity element of ${\langle}Z_E, +{\rangle}$, so ${\langle}Z_E, +{\rangle}$ is a monoid.

Therefore ${\langle}Z_E, +{\rangle}$ is a submonoid of ${\langle}Z, +{\rangle}$.

Example: Let $S=\{e, 0, 1\}$ with $*$ defined as follows:

\[ \begin{array}{cccc} * & e & 0 & 1\\ e & e & 0 & 1\\ 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 1\\ \end{array} \]

$U={\langle}S, *{\rangle}$ is a monoid with identity element $e$.

Let $X=\{0, 1\}$ with $*$ defined as follows:

\[ \begin{array}{ccc} * & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array} \]

$V={\langle}X, *{\rangle}$ is also a monoid with identity element 1.

Although $X{\subseteq}S$ is nonempty, $V$ is not a submonoid of $U$, because the two monoids have different identity elements.

Example: Let $V={\langle}S, *{\rangle}$ be a monoid with identity $e$. Then ${\langle}\{e\}, *{\rangle}$ is a submonoid of $V$, and $V$ itself is also a submonoid of $V$. These two submonoids are called trivial submonoids.

Theorem: Let $U={\langle}S, *{\rangle}$ be a commutative monoid and let $X$ be the set of all idempotent elements of $S$. Then $V={\langle}X, *{\rangle}$ is a submonoid of $U$.

Homomorphisms and Isomorphisms

Definition: Let $U={\langle}R, *{\rangle}$ and $V={\langle}S, +{\rangle}$ be semigroups, and let $f:R{\rightarrow}S$ be a mapping. If for any $a, b{\in}R$: \[f(a*b)=f(a)+f(b)\] then $f$ is called a semigroup homomorphism from $U$ to $V$, and:

If $f$ is surjective, $f$ is called a semigroup epimorphism from $U$ to $V$.
If $f$ is injective, $f$ is called a semigroup monomorphism from $U$ to $V$.
If $f$ is bijective, $f$ is called a semigroup isomorphism from $U$ to $V$.

Theorem: Let $U={\langle}X, *{\rangle}$ and $V={\langle}S, +{\rangle}$ both be semigroups and let $f$ be a semigroup homomorphism from $U$ to $V$. For any $a{\in}X$, if $a$ is an idempotent element of $U$, then $f(a)$ is an idempotent element of $V$.

Proof: Since $a{\in}X$, we have $f(a){\in}S$.

Since $a$ is an idempotent in $U$ and $f$ is a semigroup homomorphism from $U$ to $V$, then: \[f(a)=f(a^2)=f(a*a)=f(a)+f(a)=f^2(a)\] This shows that $f(a)$ is an idempotent element of $V$ with respect to $+$.

Definition: Let $U={\langle}S, *, e_*{\rangle}$ and $V={\langle}X, +, e_+{\rangle}$ be monoids, and let $f:S{\rightarrow}X$ be a mapping from $U$ to $V$. If for any $a, b{\in}S$:

$f(a*b)=f(a)+f(b)$
$f(e_*)=e_+$

then $f$ is called a monoid homomorphism. Moreover:

If $f$ is surjective, then $f$ is called a monoid epimorphism from $U$ to $V$.
If $f$ is injective, then $f$ is called a monoid monomorphism from $U$ to $V$.
If $f$ is bijective, then $f$ is called a monoid isomorphism from $U$ to $V$.

Example: Given two monoids $U={\langle}R, +, 0{\rangle}$ and $V={\langle}R^+, {\times}, 1{\rangle}$.

Define $f:R{\rightarrow}R^+$ as follows: for any $x{\in}R$, \[f(x)=5^x\]

Prove: $f$ is a monoid monomorphism from $U$ to $V$.

Proof: For any $x, y{\in}R$, we have \[f(x+y)=5^{x+y}=5^x{\times}5^y=f(x){\times}f(y)\] \[f(0)=5^0=1\] Therefore $f$ is a homomorphism from $U$ to $V$.

Also, $f$ is strictly monotonically increasing on $R$, so $f$ is injective.

Therefore, $f$ is a monoid monomorphism from $U$ to $V$.

Example: Given two monoids $U={\langle}N, +, 0{\rangle}$ and $V={\langle}S, *, e{\rangle}$, where $N$ is the set of natural numbers, $S=\{e, 0, 1\}$, $+$ is ordinary addition, and $*$ is defined as follows:

\[ \begin{array}{cccc} * & e & 0 & 1\\ e & e & 0 & 1\\ 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 1\\ \end{array} \]

Define the mapping $f:N{\rightarrow}S$, for any $m{\in}N$:

\[ f(m)=\begin{cases}0&\text{when }m{\neq}0\\1&\text{when }m=0\end{cases} \]

Is $f$ a monoid homomorphism?

Solution: For any $a, b{\in}N$,

If $a{\neq}0$ or $b{\neq}0$:

$f(a+b)=0, \, f(a)*f(b)=0$,

Thus: $f(a+b)=f(a)*f(b)$.

If $a=0, b=0$:

$f(a+b)=1, \, f(a)*f(b)=1*1=1$,

Thus: $f(a+b)=f(a)*f(b)$.

Therefore, $f$ is a semigroup homomorphism from $U$ to $V$, but since

$f(0)=1{\neq}e$

$f$ is not a monoid homomorphism from $U={\langle}N, +, 0{\rangle}$ to $V={\langle}S, *, e{\rangle}$.

Theorem: Let $f$ be a surjective homomorphism from algebraic system $U={\langle}S, +{\rangle}$ to algebraic system $V={\langle}X, *{\rangle}$, where $*$ and $+$ are both binary operations. Then:

If $U$ is a semigroup, then $V$ is also a semigroup.
If $U$ is a monoid, then $V$ is also a monoid.

Theorem: Let $U={\langle}R, *{\rangle}$, $V={\langle}S, {\otimes}{\rangle}$, and $W={\langle}T, {\oplus}{\rangle}$ all be semigroups, and let $f:R{\rightarrow}S$ and $g:S{\rightarrow}T$ be semigroup homomorphisms from $U$ to $V$ and from $V$ to $W$, respectively. Then $f{\cdot}g$ is a semigroup homomorphism from $U$ to $W$.

Proof: For any $a, b{\in}R$:

\[(f{\cdot}g)(a*b)=g(f(a*b))\]

\[=g(f(a){\otimes}f(b))\]

\[=g(f(a)){\oplus}g(f(b))\]

\[=(f{\cdot}g)(a){\oplus}(g{\cdot}f)(b)\] Therefore, $f{\cdot}g$ is a semigroup homomorphism from $U$ to $W$.

Groups and Subgroups

Groups

Definition: Given an algebraic system $U={\langle}S, *{\rangle}$, where $S$ is a nonempty set and $*$ is a binary operation on $S$. If:

$*$ is associative;
there exists an identity element $e$;
every element has an inverse.

Then $U$ is called a group.

Definition: Let $U={\langle}S, *{\rangle}$ be a monoid. If every element of $S$ is invertible, then $U$ is called a group.

For a group $U={\langle}S, *{\rangle}$, we may refer to it simply as group $S$.

Example: The algebraic system $U={\langle}R^*, {\times}{\rangle}$ is a group, where $R^*$ is the set of nonzero real numbers and ${\times}$ is ordinary multiplication.

Proof: (1) Associativity. For ${\forall}a, b, c{\in}R^*$, \[a{\times}(b{\times}c)=(a{\times}b){\times}c\] the associative law holds.

Identity element.$e=1$.
Inverse element. For ${\forall}a{\in}R^*$, $a^{-1}=1/a$,

$a{\times}1/a=1/a{\times}a=1$

Therefore, $U$ is a group.

Theorem: A group contains no zero element.

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

Theorem: Let $U={\langle}S, *{\rangle}$ be a group. Then for any $a, b{\in}S$:

There exists a unique element $x{\in}S$ such that $a*x=b$.
There exists a unique element $y{\in}S$ such that $y*a=b$.

Theorem: Let $U={\langle}S, *{\rangle}$ be a group. Then for all $a, b{\in}S$, $(a*b)^{-1}=b^{-1}*a^{-1}$.

Proof:

$(a*b)*(b^{-1}*a^{-1})$

$=a*(b*b^{-1})*a^{-1}$

$=a*e*a^{-1}$

$=a*a^{-1}=e$

$(b^{-1}*a^{-1})*(a*b)$

$=b^{-1}*(a^{-1}*a)*b$

$=b^{-1}*e*b$

$=b^{-1}*b=e$

This shows that the inverse of $a*b$ is $b^{-1}*a^{-1}$, therefore:

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

Theorem: Let $U={\langle}S, *{\rangle}$ be a group. For ${\forall}a, b, c{\in}S$:

If $a*c=b*c$, then $a=b$.
If $c*a=c*b$, then $a=b$.

That is, the operation in a group satisfies the cancellation law.

Definition: Let $U={\langle}S, *{\rangle}$ be a group.

If $S$ is finite, then $U$ is called a finite group. If $|S|=n$, then $U$ is called a group of order n. We use $|U|$ to denote the order of $U$.
If $S$ is infinite, then $U$ is called an infinite group.

Example: ${\langle}Z, +{\rangle}$ is an infinite group.

Example: ${\langle}N_k, +_k{\rangle}$ is a finite group (of order $k$), where: \[N_k=\{0, 1, 2, {\cdots}, k-1\}\] $+_k$ is addition modulo $k$.

Example: ${\langle}\{e, a\}, *{\rangle}$ is a group of order 2.

with $*$ defined as:

\[ \begin{array}{ccc} * & e & a \\ e & e & a \\ a & a & e \\ \end{array} \]

Definition: Let $a$ be an element of a group ${\langle}S, *{\rangle}$.

If there exists a positive integer $n$ such that $a^n=e$, then the smallest such positive integer is called the period (order) of $a$. We say $a$ has finite order and denote the order by $|a|$.
If no such positive integer $n$ exists, then $a$ is said to have infinite order.

Theorem: Let $U={\langle}S, *{\rangle}$ be a group, $a{\in}S$ be an element of order $n$, and $k$ be an integer. Then:

$a^k=e$ if and only if $k$ is a multiple of $n$;
the order of $a^{-1}$ equals the order of $a$;
if $U$ is a finite group, then the order of every element is at most $|S|$.

Example: Let $G={\langle}K_4, *{\rangle}$ be a group with $K_4=\{a, b, c, d\}$, and $*$ defined as follows:

\[ \begin{array}{ccccc} * & a & b & c & d\\ a & a & b & c & d\\ b & b & a & d & c\\ c & c & d & a & b\\ d & d & c & b & a\\ \end{array} \]

Find:

The order of each element;
$|G|$.

Solution:

The identity element is $e=a$.

$a^1=a$, $b^2=a$, $c^2=a$, $d^2=a$

Therefore, every non-identity element has order 2.

\[|G|=4.\]

Subgroups

Definition: Let $U={\langle}G, *{\rangle}$ be a group and let $S{\subseteq}G$ be a nonempty subset. If $V={\langle}S, *{\rangle}$ also forms a group, then $V$ is called a subgroup of $U$. If $S$ is a proper subset of $G$, $V$ is called a proper subgroup. If $S=\{e\}$ or $S=G$, $V$ is called a trivial subgroup.

Theorem: Let $U={\langle}G, *{\rangle}$ be a group and let $S{\subseteq}G$ be a nonempty subset. $V={\langle}S, *{\rangle}$ is a subgroup of $U$ if and only if:

For ${\forall}a, b{\in}S$, $a*b{\in}S$.
$V$ has an identity element.
For ${\forall}a{\in}S$, $a$ has an inverse $a^{-1}{\in}S$.

For example, the group ${\langle}Z, +{\rangle}$ is a subgroup of the group ${\langle}R, +{\rangle}$.

Let ${\langle}Z, +{\rangle}$ be a group and $Z_E=\{x|x=2n, n{\in}Z\}$. Prove: ${\langle}Z_E, +{\rangle}$ is a subgroup of ${\langle}Z, +{\rangle}$.

Proof: Clearly $Z_E{\subseteq}Z$ is a nonempty subset.

For ${\forall}x, y{\in}Z_E$, let $x=2n$, $y=2m$, so $n, m{\in}Z$, $n+m{\in}Z$. $x+y=2n+2m=2(n+m){\in}Z_E$, so closure holds.
The identity element is 0. For ${\forall}x{\in}Z_E$, write $x=2n$: $0+2n=2n+0=2n$.
For ${\forall}x{\in}Z_E$, let $x=2n$, so $n{\in}Z$.

Then $-x=-2n=2(-n)$, and since $-n{\in}Z$, we have $2(-n){\in}Z_E$, i.e., $-x{\in}Z_E$.

And $x+(-x)=(-x)+x=0$. Therefore the inverse of $x$ is $-x$, so every element has an inverse.

Therefore ${\langle}Z_E, +{\rangle}$ is a subgroup of ${\langle}Z, +{\rangle}$.

Theorem: The identity element of a subgroup is the same as that of the group.

Proof: Let ${\langle}S, *{\rangle}$ be any subgroup of ${\langle}G, *{\rangle}$, and let $e_1$ and $e$ be the identity elements of $S$ and $G$ respectively. For any $x{\in}S{\subseteq}G$:

\[e_1*x=x,\,\, e*x=x\] Therefore: $e_1*x=e*x$

By the cancellation law: $e_1=e$.

Theorem: Let $U={\langle}G, *{\rangle}$ be a group and let $S{\subseteq}G$ be nonempty. Then $V={\langle}S, *{\rangle}$ is a subgroup of $U$ if and only if for any $a, b{\in}S$, $a*b^{-1}{\in}S$.

If $U_1={\langle}S_1, *{\rangle}$ and $U_2={\langle}S_2, *{\rangle}$ are both subgroups of $U={\langle}S, *{\rangle}$, then $U_3={\langle}S_1{\cap}S_2, *{\rangle}$ is also a subgroup of $U$.

Proof: Let $e$ be the identity of $U$. Then $e{\in}S_1$, $e{\in}S_2$, so $e{\in}S_1{\cap}S_2$, thus $S_1{\cap}S_2$ is a nonempty subset of $S$.

For any $a, b{\in}S_1{\cap}S_2$, we have $a, b{\in}S_1$ and $a, b{\in}S_2$. Since $U_1$ is a subgroup of $U$, $a*b^{-1}{\in}S_1$.

Since $U_2$ is a subgroup of $U$: $a*b^{-1}{\in}S_2$. Therefore $a*b^{-1}{\in}S_1{\cap}S_2$, so $U_3$ is a subgroup of $U$.

Theorem: Let $U={\langle}G, *{\rangle}$ be a group and let $S$ be a nonempty subset of $G$. If $S$ is finite, then $V={\langle}S, *{\rangle}$ is a subgroup of $U$ if and only if: for ${\forall}a, b{\in}S$, $a*b{\in}S$.

Special Groups

Abelian Groups

Definition: If the operation $*$ in a group $U={\langle}S, *{\rangle}$ satisfies the commutative law, then $U$ is called an Abelian group (commutative group). Otherwise, $U$ is called a non-Abelian group.

Example: ${\langle}R, +{\rangle}$, ${\langle}Z, +{\rangle}$, ${\langle}R-\{0\}, {\times}{\rangle}$, and ${\langle}Q-\{0\}, {\times}{\rangle}$ are all Abelian groups, where $Q$, $R$, and $Z$ are the sets of rational, real, and integer numbers, and $+$ and ${\times}$ are ordinary addition and multiplication.

Example: Let $G$ be the set of all nonsingular matrices of order $n$, with matrix multiplication “${\cdot}$” as the binary operation defined on $G$. ***

Prove: The algebraic system ${\langle}G, {\cdot}{\rangle}$ is a non-Abelian group.

Proof:

For ${\forall}A, B, C{\in}G$: $A{\cdot}(B{\cdot}C)=(A{\cdot}B){\cdot}C$, so $\cdot$ is associative.
The identity element of ${\langle}G, {\cdot}{\rangle}$ is the identity matrix $E$. Because for ${\forall}A{\in}G$: $E{\cdot}A=A{\cdot}E=A$.
For ${\forall}A{\in}G$, $A$ has a unique inverse, namely the inverse matrix $A^{-1}$, because:

$A^{-1}{\cdot}A=A{\cdot}A^{-1}=E.$

Therefore ${\langle}G, {\cdot}{\rangle}$ is a group. However, for ${\forall}A, B{\in}G$, $A{\cdot}B{\neq}B{\cdot}A$, so ${\langle}G, {\cdot}{\rangle}$ is a non-Abelian group.

Example: Let $G={\langle}K_4, *{\rangle}$ be a group with $K_4=\{a, b, c, d\}$, and $*$ defined as follows.

\[ \begin{array}{ccccc} * & a & b & c & d\\ a & a & b & c & d\\ b & b & a & d & c\\ c & c & d & a & b\\ d & d & c & b & a\\ \end{array} \]

Prove: ${\langle}K_4, *{\rangle}$ is an Abelian group.

Proof: From the operation table, the table is symmetric, so $*$ satisfies the commutative law, thus ${\langle}K_4, *{\rangle}$ is an Abelian group.

Theorem: Let $U={\langle}S, *{\rangle}$ be a group. Then $U$ is an Abelian group if and only if for any $a, b{\in}S$: $(a*b)^2=a^2*b^2$.

Proof: Sufficiency. For any $a, b{\in}S$: $(a*b)^2 = a^2*b^2$, That is:

$(a*b)*(a*b) = (a*a)*(b*b)$.

And $a*(b*a)*b=(a*b)*(a*b)$

$=(a*a)*(b*b)=a*(a*b)*b$.

By cancellation: $b*a=a*b$, so $U$ is an Abelian group.

Necessity. Since $U$ is an Abelian group, for any $a, b{\in}S$: \[a*b=b*a\]

Therefore \[(a*b)^2=(a*b)*(a*b)\] \[=a*(b*a)*b\] \[=a*(a*b)*b\] \[=(a*a)*(b*b)\] \[=a^2*b^2\] This completes the proof.

Cyclic Groups

Definition: Let $U={\langle}S, *{\rangle}$ be a group. If there exists an element $g{\in}S$ such that every element of $S$ can be expressed as a power of $g$, i.e., $S=\{g^n|n{\in}Z\}$, then $U$ is called a cyclic group.

$ is called the generator of $U$, and $U$ can be written as $U={\langle}{\langle}g{\rangle}, *{\rangle}$.

Proof: (1) Modular arithmetic satisfies the associative law, i.e., for ${\forall}a, b, c{\in}X$: $(a{\oplus}b){\oplus}c=a{\oplus}(b{\oplus}c)$. Therefore ${\langle}X, {\oplus}{\rangle}$ is a semigroup.

The identity element is 0, so ${\langle}X, {\oplus}{\rangle}$ is a monoid.
Inverse elements.

$60^{-1}=300$, $120^{-1}=240$, $180^{-1}=180$, $240^{-1}=120$, $300^{-1}=60$.

Therefore ${\langle}X, {\oplus}{\rangle}$ is a group.

Generator $g=60$. Because:

$g^0=0$, $g^1=60$, $g^2=120$, $g^3=180$, $g^4=240$, $g^5=300$, $g^6=0$, $g^7=60$, ${\cdots}$

So ${\langle}X, {\oplus}{\rangle}$ is a cyclic group.

In fact, this cyclic group has another generator $g=300$. Because:

$g^0=0$, $g^1=300$, $g^2=240$, $g^3=180$, $g^4=120$, $g^5=60$, $g^6=0$, $g^7=300$, ${\cdots}$

For a cyclic group $U={\langle}{\langle}g{\rangle}, *{\rangle}$, based on the order of the generator, there are two types:

When the order of the generator $g$ is infinite, $U$ is an infinite cyclic group, and: \[{\langle}g{\rangle}=\{g^n|n{\in}Z, \text{when }m{\neq}n, g^m{\neq}g^n\}\]
When the order of the generator $g$ is finite, $U$ is a finite cyclic group. If $|g|=n$, then: \[{\langle}g{\rangle}=\{e, g, g^2, {\cdots}, g^{n-1}\}.\]

Example: Let $X=\{0, 60, 120, 180, 240, 300\}$, ${\oplus}$ be a binary operation on $X$, for ${\forall}a, b{\in}X$: $a{\oplus}b=(a+b)\, \mod\, 360$. ${\langle}X, {\oplus}{\rangle}$ is a cyclic group of order 6.

Because the generator is $g=60$. $g^0=0$, $g^1=60$, $g^2=120$, $g^3=180$, $g^4=240$, $g^5=300$, $g^6=0$, $g^7=60$, ${\cdots}$, Therefore, $|g|=6$, $X={\langle}g{\rangle}=\{g^0, g^1, g^2, g^3, g^4, g^5\}.$

Concerning the number of generators of a cyclic group $U={\langle}{\langle}g{\rangle}, *{\rangle}$, we have the following theorem.

Theorem:

If $U$ is an infinite cyclic group, then $U$ has exactly two generators: $g$ and $g^{-1}$.
If $U$ is a finite cyclic group of order $n$, then $U$ has $\varphi(n)$ generators.

$\varphi(n)$ is Euler’s totient function, representing the count of positive integers less than $n$ that are coprime to $n$.

Example: Euler’s totient of 9: $\varphi(9)=6$.

Because the positive integers less than 9 and coprime to 9 are 1, 2, 4, 5, 7, 8, totalling 6.

For example, let $X=\{0, 60, 120, 180, 240$,$300\}, {\oplus}$ be a binary operation on $X$, for ${\forall}a, b{\in}X$:

\[a{\oplus}b=(a+b)\, \mod\, 360.\]

The cyclic group ${\langle}X, {\oplus}{\rangle}$ of order 6 has $\varphi(6)=2$ generators: 60 and 300.

Example: The integer additive group $U={\langle}Z, +{\rangle}$ is an infinite cyclic group.

Solution: The identity of $U$ is 0; for any $x{\in}Z$, the inverse of $x$ is $-x$, i.e., $x^{-1}=-x$. Both 1 and -1 are generators of $U$.

For generator 1, for any positive integer $n$:

$1^0=0$, $1^1=1$, $1^2=1+1=2$, ${\cdots}$, $1^n=1+1+{\cdots}+1=n$, ${\cdots}$.

$1^{-1}=-1$, $1^{-2}=(1^{-1})^2=(-1)^2=(-1)+(-1)=-2$, ${\cdots}$, $1^{-n}=(1^{-1})^n=(-1)^n=-n$, ${\cdots}$.

$n=(1+1+\cdots+1)=1^n$,

$-n=(-1)+(-1)+\cdots+(-1)=(-1)^n$.

Example: For the cyclic group of order 12: $G={\langle}{\langle}g{\rangle}, *{\rangle}={\langle}\{e, g, g^2, {\cdots}, g^{11}\}, *{\rangle}$, the numbers less than or equal to 12 that are coprime to 12 are 1, 5, 7, 11, so $\varphi(12)=4$. $ has four generators: $g, g^5, g^7, g^{11}$.

Theorem: Let $U={\langle}{\langle}g{\rangle}, *{\rangle}$ be a cyclic group:

If $g$ has finite order ($|g|=m$), then ${\langle}{\langle}g{\rangle}, *{\rangle}$ is isomorphic to ${\langle}N_m, +_m{\rangle}$.
If $g$ has infinite order, then ${\langle}{\langle}g{\rangle}, *{\rangle}$ is isomorphic to ${\langle}Z, +{\rangle}$.

Theorem: Let $U={\langle}S, *{\rangle}$ be a finite cyclic group generated by $g$. If $|S|=n$, then $g^n=e$ and: \[S=\{g^1, g^2, {\cdots}, g^n=g^0=e\}\] where $n$ is the smallest positive integer such that $g^n=e$.

Theorem: Every cyclic group is an Abelian group.

Proof: Let $U={\langle}S, *{\rangle}$ be a cyclic group with generator $g$. For any $x, y{\in}S$, there exist integers $m, n{\in}Z$ such that: $x=g^m, y=g^n$. Then: $x*y=g^m*g^n=g^{m+n},\,\,y*x=g^n*g^m=g^{n+m}=g^{m+n}$. Therefore $x*y=y*x$, so $U$ is an Abelian group.

Symmetric Groups and Permutation Groups

Definition: Let $S$ be a nonempty finite set. A bijection (one-to-one mapping) on $S$ is called a permutation on $S$; if $S$ has $n$ elements, it is called an $-permutation. If $|S|=n$, let $S_n$ denote the set of all $n$-permutations on $S$. \[S_n=\{f\mid f:S{\rightarrow}S \text{ is a permutation}\}\]

The composition of permutations is defined as: for ${\forall}x{\in}S$, $f{\circ}g(x)=g(f(x))$; $f{\circ}g$ is called the permutation product of $f$ and $g$.

For any two permutations $f, g$ on $S$, the composition $f{\circ}g$ is also a permutation on $S$, so permutation composition is closed.

Such composition satisfies the associative law, and there exists an identity permutation $E$ (identity element) on $S$ such that for any permutation $f$ on $S$: $E{\circ}f=f{\circ}E=f$

Moreover, for each permutation $f$, there exists a permutation $f^{-1}$ on $S$ such that: $f^{-1}{\circ}f=f{\circ}f^{-1}=E$.

This shows that the set $S_n$ of all permutations on $S$ with composition ${\langle}S_n, \circ{\rangle}$ forms a group.

Definition: Let $S_n$ be the set of all permutations on $S$. Then ${\langle}S_n, {\circ}{\rangle}$ forms a group called the symmetric group of degree $n$ on $S$.

Any subgroup of ${\langle}S_n, {\circ}{\rangle}$ is called a permutation group of degree $n$ on $S$.

Example: Let $S=\{1, 2, 3\}$. The symmetric group of degree 3 on $S$ is ${\langle}S_3, \circ{\rangle}$, where:

$S_3=\{{\pi}_e, {\pi}_1, {\pi}_2, {\pi}_3, {\pi}_4, {\pi}_5\}$

\[ {\pi}_e=\left(\begin{matrix} 1 & 2 & 3\\ 1 & 2 & 3\\ \end{matrix}\right) ,\,\, {\pi}_1=\left(\begin{matrix} 1 & 2 & 3\\ 2 & 1 & 3\\ \end{matrix}\right) \] \[ {\pi}_2=\left(\begin{matrix} 1 & 2 & 3\\ 3 & 2 & 1\\ \end{matrix}\right) ,\,\, {\pi}_3=\left(\begin{matrix} 1 & 2 & 3\\ 1 & 3 & 2\\ \end{matrix}\right) \]

\[ {\pi}_4=\left(\begin{matrix} 1 & 2 & 3\\ 2 & 3 & 1\\ \end{matrix}\right) ,\,\, {\pi}_5=\left(\begin{matrix} 1 & 2 & 3\\ 3 & 1 & 2\\ \end{matrix}\right) \]

\[ \begin{array}{ccccccc} \circ & {\pi}_e & {\pi}_1 & {\pi}_2 & {\pi}_3 & {\pi}_4 & {\pi}_5\\ {\pi}_e & {\pi}_e & {\pi}_1 & {\pi}_2 & {\pi}_3 & {\pi}_4 & {\pi}_5\\ {\pi}_1 & {\pi}_1 & {\pi}_e & {\pi}_5 & {\pi}_4 & {\pi}_3 & {\pi}_2\\ {\pi}_2 & {\pi}_2 & {\pi}_4 & {\pi}_e & {\pi}_5 & {\pi}_1 & {\pi}_3\\ {\pi}_3 & {\pi}_3 & {\pi}_5 & {\pi}_4 & {\pi}_e & {\pi}_1 & {\pi}_1\\ {\pi}_4 & {\pi}_4 & {\pi}_2 & {\pi}_3 & {\pi}_1 & {\pi}_5 & {\pi}_e\\ {\pi}_5 & {\pi}_5 & {\pi}_3 & {\pi}_1 & {\pi}_2 & {\pi}_e & {\pi}_4\\ \end{array} \]

We can verify: ${\langle}S_3, {\circ}{\rangle}$ is the symmetric group of degree 3 (order 6),

${\langle}\{{\pi}_e\}, {\circ}{\rangle}$ is a permutation group of degree 3 (order 1),

${\langle}\{{\pi}_e, {\pi}_1\}, {\circ}{\rangle}$, ${\langle}\{{\pi}_e, {\pi}_2\}, {\circ}{\rangle}$ and ${\langle}\{{\pi}_e, {\pi}_3\}$, {}{}$ are both second-order ternary permutation groups,

${\langle}\{{\pi}_e, {\pi}_4, {\pi}_5\}, {\circ}{\rangle}$ is a permutation group of degree 3 (order 3).

Definition: Let $U={\langle}G, {\circ}{\rangle}$ be a permutation group on set $S$. The binary relation: \[R=\{{\langle}a, b{\rangle}|{\pi}(a)=b, {\pi}{\in}G\}\] is called the binary relation on $S$ induced by $U$.

Example: Let $S=\{1, 2, 3\}$ and $G=\{{\pi}_e, {\pi}_1\}$.

\[ {\pi}_e=\left(\begin{matrix} 1 & 2 & 3\\ 1 & 2 & 3\\ \end{matrix}\right) ,\,\, {\pi}_1=\left(\begin{matrix} 1 & 2 & 3\\ 2 & 1 & 3\\ \end{matrix}\right) \]

G, {}{}$ is a permutation group of degree 3 (order 2) on $S$, and the binary relation it induces is:

\[R=\{{\langle}1, 1{\rangle}, {\langle}2, 2{\rangle}, {\langle}3, 3{\rangle}, {\langle}1, 2{\rangle}, {\langle}2, 1{\rangle}\}\]

Theorem: The binary relation on $S$ induced by a permutation group ${\langle}G, \circ{\rangle}$ is an equivalence relation.

In the above example, clearly:

$R=\{{\langle}1, 1{\rangle}, {\langle}2, 2{\rangle}, {\langle}3, 3{\rangle}, {\langle}1, 2{\rangle}, {\langle}2, 1{\rangle}\}$

is reflexive, symmetric, and transitive, hence is an equivalence relation.

Definition: Let $U={\langle}G, *{\rangle}$ and $V={\langle}H, \circ{\rangle}$ be groups, and let $f:G{\rightarrow}H$ be a mapping. If for any $a, b{\in}G$: \[f(a*b)=f(a){\circ}f(b)\] then $f$ is called a group homomorphism from $U$ to $V$, and:

If $f$ is surjective, then $f$ is called a group epimorphism from $U$ to $V$.
If $f$ is injective, then $f$ is called a group monomorphism from $U$ to $V$.
If $f$ is bijective, then $f$ is called a group isomorphism from $U$ to $V$.

In particular, if $f$ is a homomorphism (isomorphism) from group $U$ to itself, then $f$ is called an endomorphism (automorphism) of $U$.

Example: Let group $G_1={\langle}M_n^*(R), *{\rangle}$ and group $G_2={\langle}R^*, {\times}{\rangle}$.

where $R^*$ is the set of nonzero real numbers, $M_n^*(R)$ is the set of real invertible $n{\times}n$ matrices, and the operations $*$ and ${\times}$ in $G_1$ and $G_2$ are matrix multiplication and scalar multiplication, respectively.

The function $f:M_n^*(R){\rightarrow}R^*$ is defined as: for any $A{\in}M_n^*(R)$: \[f(A)=|A|\]

Prove: $f$ is a group epimorphism from $G_1$ to $G_2$.

Proof: For any $A, B{\in}M_n^*(R)$: \[f(A*B)=|A*B|=|A|{\times}|B|=f(A){\times}f(B)\] Therefore $f$ is a group homomorphism.

Moreover, for any $r{\in}R^*$, take

\[ A=\left( \begin{matrix} r & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots & \cdots & \vdots\\ 0 & 0 & \cdots & 1\\ \end{matrix} \right){\in}M_n^*(R) \]

Then: $f(A)=|A|=r$. So $f$ is surjective, thus $f$ is a group epimorphism from $G_1$ to $G_2$.

Theorem: Let $f$ be a homomorphism from group $U={\langle}G, *{\rangle}$ to group $V={\langle}H, \circ{\rangle}$. $e_G$ and $e_H$ are the identity elements of $U$ and $V$ respectively, then:

$f(e_G)=e_H$
$f(x^{-1})=f^{-1}(x)$ (${\forall}x{\in}G$)

Rings

Definition: An algebraic system $U={\langle}S, +, *{\rangle}$ with two binary operations satisfies:

${\langle}S, +{\rangle}$ is an Abelian group;
${\langle}S, *{\rangle}$ is a semigroup;
The multiplication $*$ satisfies the distributive law over addition $+$, i.e., for any $a$, $b$, $c{\in}S$:

$a*(b+c)=a*b+a*c$, $(b+c)*a=b*a+c*a$, then $U$ is called a ring.

Example: ${\langle}Z, +, {\times}{\rangle}$, ${\langle}Q, +, {\times}{\rangle}$, ${\langle}R, +, {\times}{\rangle}$, ${\langle}C, +, {\times}{\rangle}$ are all rings, called the ring of integers, ring of rationals, ring of reals, and ring of complex numbers, respectively. These rings are collectively called number rings.

Example: For ${\langle}Z, +, {\times}{\rangle}$, since:

${\langle}Z, +{\rangle}$ is an Abelian group;
${\langle}Z, {\times}{\rangle}$ is a semigroup;
Multiplication ${\times}$ satisfies the distributive law over addition $+$:

${\forall}a, b, c{\in}S$, we have: $a{\times}(b+c)=a{\times}b+a{\times}c, \, \, (b+c){\times}a=b{\times}a+c{\times}a$.

Therefore ${\langle}Z, +, {\times}{\rangle}$ is a ring.

In a ring ${\langle}S, +, *{\rangle}$, the identity element of the addition operation is denoted by 0, called the zero element of the ring; the identity element of the multiplication operation (if it exists) is denoted by 1, called the unity element (identity element) of the ring.

The inverse of an element $a$ with respect to addition in a ring is denoted $-a$, called the negative element of $a$.

$a+(-b)$ is usually written as $a-b$. If the inverse with respect to multiplication exists, it is denoted $a^{-1}$, called the inverse element of $a$.

For brevity, $a*b$ may be written as $ab$. The $n$-th powers of $a$ under $+$ and $*$ are written as $na$ and $a^n$ respectively, that is: \[na=a+a+\cdots+a\] \[a^n=a*a*\cdots*a\]

By convention, without parentheses, the order of operations is: exponentiation before multiplication, multiplication before addition.

Definition: In a ring $R={\langle}S, +, *{\rangle}$, if the operation $*$ is commutative, then $R$ is called a commutative ring.

For example, all number rings are commutative rings.

Prove: The algebraic system ${\langle}Z, {\oplus}, {\odot}{\rangle}$ is a commutative ring, where ${\oplus}$ and ${\odot}$ are defined as follows: ${\forall}a, b{\in}Z$, $a{\oplus}b=a+b-1$, $a{\odot}b=a+b-ab$.

Proof: (1) ${\langle}Z, {\oplus}{\rangle}$ is an Abelian group.

Associativity. ${\forall}a, b, c{\in}Z$, \[(a{\oplus}b){\oplus}c=(a+b-1){\oplus}c=(a+b-1)+c-1=a+b+c-2\] \[a{\oplus}(b{\oplus}c)=a{\oplus}(b+c-1)=a+(b+c-1)-1=a+b+c-2\] \[(a{\oplus}b){\oplus}c=a{\oplus}(b{\oplus}c)\]

Identity element (zero element).

1 is the identity element. Because ${\forall}a{\in}Z$,

\[1{\oplus}a=1+a-1=a\] \[a{\oplus}1=a+1-1=a\] \[1{\oplus}a=a{\oplus}1=a\]

Inverse element (negative element).

For ${\forall}a{\in}Z$, the inverse of $a$ is $2-a$. Because: \[a{\oplus}(2-a)=a+2-a-1=1\] \[(2-a){\oplus}a=2-a+a-1=1\] \[a{\oplus}(2-a)=(2-a){\oplus}a=1\]

Commutativity.

${\forall}a, b{\in}Z$, \[a{\oplus}b=a+b-1=b+a-1=b{\oplus}a\]

Therefore: ${\langle}Z, {\oplus}{\rangle}$ is an Abelian group.

${\langle}Z, {\odot}{\rangle}$ is a semigroup.

Associativity.

${\forall}a, b, c{\in}Z$, \[(a{\odot}b){\odot}c=(a+b-ab){\odot}c\] \[=(a+b-ab)+c-(a+b-ab)c\] \[=a+b+c-ab-ac-bc+abc\]

\[a{\odot}(b{\odot}c)=a{\odot}(b+c-bc)\] \[=a+(b+c-bc)-a(b+c-bc)\] \[=a+b+c-ab-ac-bc+abc\] \[(a{\odot}b){\odot}c=a{\odot}(b{\odot}c)\] Therefore: ${\langle}Z, {\odot}{\rangle}$ is a semigroup.

Distributive law of ${\odot}$ over ${\oplus}$. ${\forall}a, b, c{\in}Z$, \[a{\odot}(b{\oplus}c)=a{\odot}(b+c-1)\] \[=a+b+c-1-a(b+c-1)\] \[=a+b-ab+a+c-ac-1\] \[=a{\odot}b+a{\odot}c-1=(a{\odot}b){\oplus}(a{\odot}c)\]

\[(b{\oplus}c){\odot}a=(b+c-1){\odot}a\] \[=b+c-1+a-(b+c-1)a\] \[=b+c-1+a-ba-ca+a\] \[=b+a-ba+c+a-ca-1\] \[=(b{\odot}a){\oplus}(c{\odot}a)\]

Therefore: ${\langle}Z, {\oplus}, {\odot}{\rangle}$ is a ring.

Commutativity of multiplication.

${\forall}a, b{\in}Z$, \[a{\odot}b=a+b-ab\] \[b{\odot}a=b+a-ba=a+b-ab\] Thus \[a{\odot}b=b{\odot}a\]

Therefore: ${\langle}Z, {\oplus}, {\odot}{\rangle}$ is a commutative ring.

Definition: In a ring $U={\langle}S, +, *{\rangle}$, if ${\langle}S, *{\rangle}$ is a monoid, then $U$ is called a unital ring (ring with unity).

For example, the rings of integers, rationals, and reals are all unital rings (with unity 1), and the ring of complex numbers also has unity 1 (i.e., $1+0i$, where $i$ is the imaginary unit).

For ${\langle}Z, +, {\times}{\rangle}$, since ${\langle}Z, +{\rangle}$ is an Abelian group, ${\langle}Z, {\times}{\rangle}$ is a monoid, and ${\times}$ satisfies the distributive law over $+$, therefore ${\langle}Z, +, {\times}{\rangle}$ is a unital ring.

Another example: let $E_V$ be the set of even integers, i.e., $E_V=\{2i|i{\in}Z\}$, with $+$ and ${\times}$ as ordinary addition and multiplication. Then $U={\langle}E_V, +, {\times}{\rangle}$ forms a ring, but this ring has no unity element, so $U$ is not a unital ring.

Example: Let ${\langle}A, +{\rangle}$ be an Abelian group. Define the operation $*$ on $A$ by: for ${\forall}a, b{\in}A$, $a*b=0$.

Here 0 is the identity element of ${\langle}A, +{\rangle}$; then ${\langle}A, +, *{\rangle}$ is a ring.

Proof: It is given that ${\langle}A, +{\rangle}$ is an Abelian group.

Prove ${\langle}A, *{\rangle}$ is a semigroup. For ${\forall}a, b, c{\in}A$:

$a*b=0{\in}A$, so closure holds.

Moreover $a*(b*c)=a*0=0$, $(a*b)*c=0*c=0$, So $a*(b*c)=(a*b)*c$, thus associativity holds.

Therefore ${\langle}A, *{\rangle}$ is a semigroup.

Distributive law of $*$ over $+$. For ${\forall}a, b, c{\in}A$, $b+c{\in}A$$

$a*(b+c)=0$, $a*b+a*c=0+0=0$

Therefore: $a*(b+c)=a*b+a*c$

$(b+c)*a=0$, $b*a+c*a=0+0=0$

Therefore: $(b+c)*a=b*a+c*a$

$*$ satisfies the distributive law over $+$.

Therefore: ${\langle}A, +, *{\rangle}$ is a ring. This ring is called the zero ring.

Definition: Let $a$, $b$ be two nonzero elements of a ring $U={\langle}S, +, *{\rangle}$. If $a*b=0$, then $a$ is called a left zero divisor and $b$ is called a right zero divisor of $U$. If an element is both a left zero divisor and a right zero divisor, it is called a zero divisor.

Example: Let $R$ be the set of real numbers. In the set $R{\times}R$ of all ordered pairs of real numbers, define the operations ${\oplus}$ and ${\odot}$ as follows:

For any ${\langle}a_1, b_1{\rangle}$, ${\langle}a_2, b_2{\rangle}{\in}R{\times}R$

\[{\langle}a_1, b_1{\rangle}{\oplus}{\langle}a_2, b_2{\rangle}={\langle}a_1+a_2, b_1+b_2{\rangle}\] \[{\langle}a_1, b_1{\rangle}{\odot}{\langle}a_2, b_2{\rangle}={\langle}a_1{\times}a_2, b_1{\times}b_2{\rangle}\] Then ${\langle}R{\times}R, {\oplus}, {\odot}{\rangle}$ forms a ring.

The zero element is ${\langle}0, 0{\rangle}$, the unity element is ${\langle}1, 1{\rangle}$, and for any $a{\neq}0$, $b{\neq}0$. ${\langle}a, 0{\rangle}{\neq}{\langle}0, 0{\rangle},\,\,{\langle}0, b{\rangle}{\neq}{\langle}0, 0{\rangle}$ are nonzero elements. But: \[{\langle}a, 0{\rangle}{\odot}{\langle}0, b{\rangle}={\langle}0, 0{\rangle},\,\,\, {\langle}0, b{\rangle}{\odot}{\langle}a, 0{\rangle}={\langle}0, 0{\rangle}\] are all zero divisors.

Definition: A ring $U={\langle}S, +, *{\rangle}$ is called a zero-divisor-free ring if for any nonzero elements $a, b{\in}S$, we have $a*b{\neq}0$.

All number rings are zero-divisor-free rings. For example, for ${\langle}Z, +, {\times}{\rangle}$, if ${\forall}a, b{\in}Z$ and $a{\times}b=0$, then $a=0$ or $b=0$, so ${\langle}Z, +, {\times}{\rangle}$ is a zero-divisor-free ring.

By the definition of a zero-divisor-free ring, when a ring has no left zero divisors (which also means no right zero divisors), it is called a zero-divisor-free ring.

Theorem: Let $R={\langle}S, +, *{\rangle}$ be a ring. For ${\forall}a, b, c{\in}S$, we have:

$a*0=0*a=0.$
$(-a)*b=a*(-b)=-a*b.$
$(-a)*(-b)=a*b.$
$a*(b-c)=a*b-a*c$, $(b-c)*a=b*a-c*a$.

Theorem: A ring $U={\langle}S, +, *{\rangle}$ is a zero-divisor-free ring if and only if the operation $*$ satisfies the cancellation law.

That is: $U={\langle}S, +, *{\rangle}$ is a zero-divisor-free ring, for ${\forall}a, b, c{\in}S$ with $a{\neq}0$, if $a*b=a*c$ and $b*a=c*a$, then $b=c$.

Definition: If $U$ is a commutative, unital, zero-divisor-free ring, then $U$ is called an integral domain. All number rings are integral domains.

Example: Let $M_2(R)$ be the set of all $2{\times}2$ real matrices, with $+$ and $*$ being ordinary matrix addition and multiplication. Then ${\langle}M_2(R), +, *{\rangle}$ is a non-commutative ring with unity, and therefore not an integral domain.

Because the zero element and unity element of this ring are respectively:

\[ \left( \begin{matrix} 0 & 0\\ 0 & 0\\ \end{matrix} \right) \text{and} \left( \begin{matrix} 1 & 0\\ 0 & 1\\ \end{matrix} \right) \]

Obviously, matrix multiplication does not satisfy the commutative law, i.e., $A*B{\neq}B*A$, so the operation $*$ is not commutative.

Therefore, ${\langle}M_2(R), +, *{\rangle}$ is not an integral domain. Is it a zero-divisor-free ring?

Example: Let $U={\langle}S, +, *{\rangle}$ be a unital ring with at least two elements, where $\mathbf{0}$ and $\mathbf{1}$ are the zero element and unity element of $U$ respectively. Then $\mathbf{0}{\neq}\mathbf{1}$.

Proof: By contradiction. If $\mathbf{0}=\mathbf{1}$, since $S$ has at least two elements, there must be a nonzero element $a{\neq}\mathbf{0}$ in $S$.

Because: \[\mathbf{0}*a=\mathbf{0}\] \[\mathbf{0}=\mathbf{0}*a=\mathbf{1}*a=a\] Then: $\mathbf{0}=a$

This contradicts our assumption. Therefore $\mathbf{0}{\neq}\mathbf{1}$.

Definition: A ring $U={\langle}S, +, *{\rangle}$ is called a division ring if:

$S$ contains at least two elements;
$S$ contains a unity element 1;
Every element of $S-\{0\}$ has an inverse element.

Such a ring $U$ is called a division ring.

Theorem: A division ring satisfies the cancellation law and has no zero divisors.

Definition: A division ring that satisfies the commutative law is called a commutative division ring (field).

Subrings and Ring Homomorphisms

Subrings

Definition: Let $U={\langle}R, +, *{\rangle}$ be a ring and $S$ be a nonempty subset of $R$. If $V={\langle}S, +, *{\rangle}$ also forms a ring, then $V$ is called a subring of $U$, and $U$ is called an extension ring of $V$.

In particular, when $S=R$ or $S=\{0\}$, $V$ is also a subring of $U$; these two subrings are called trivial subrings.

For example, the ring of even integers ${\langle}E_V, +, {\times}{\rangle}$ is a subring of the ring of integers ${\langle}Z, +, {\times}{\rangle}$.

The rings of rationals $Q$, reals $R$, and integers $Z$ are all subrings of the ring of complex numbers $C$.

Theorem: Let $U={\langle}R, +, *{\rangle}$ be a ring and $S{\subseteq}R$ be nonempty. Then $V={\langle}S, +, *{\rangle}$ is a subring of $U$ if and only if:

${\langle}S, +{\rangle}$ is a subgroup of ${\langle}R, +{\rangle}$.
${\langle}S, *{\rangle}$ is a subsemigroup of ${\langle}R, *{\rangle}$.

Theorem: Let $U={\langle}R, +, *{\rangle}$ be a ring and $S{\subseteq}R$. Then $V={\langle}S, +, *{\rangle}$ is a subring of $U$ if and only if for any $a, b{\in}S$, we have $a-b{\in}S$ and $a*b{\in}S$.

Prove that ${\langle}2Z, +, {\times}{\rangle}$ is a subring of the ring of integers ${\langle}Z, +, {\times}{\rangle}$, where $2Z=\{2n|n{\in}Z\}$.

Proof: Clearly $2Z{\subseteq}Z$ and $2Z$ is nonempty (since $0{\in}2Z$).

For ${\forall}2m, 2n{\in}2Z$, we have $m, n{\in}Z$, $m-n{\in}Z$, $-n{\in}Z$, $-2n=2(-n){\in}2Z$.

Therefore: $2m-2n=2(m-n){\in}2Z,\,\,2m{\times}2n=2(m{\times}2n){\in}2Z$ (since $2n{\in}Z$, $m{\times}2n{\in}Z$), Thus: ${\langle}2Z, +, {\times}{\rangle}$ is a subring of ${\langle}Z, +, {\times}{\rangle}$.

Ring Homomorphisms

Definition: Let $U={\langle}S, +, *{\rangle}$ and $V={\langle}X, {\oplus}, {\odot}{\rangle}$ be two rings. If there is a mapping $f:S{\rightarrow}X$ such that for any $a, b{\in}S$: $f(a+b)=f(a){\oplus}f(b),\,\,f(a*b)=f(a){\odot}f(b)$, then $f$ is called a ring homomorphism from $U$ to $V$. Moreover:

If $f$ is surjective, then $f$ is called a ring epimorphism from $U$ to $V$.
If $f$ is injective, then $f$ is called a ring monomorphism from $U$ to $V$.
If $f$ is bijective, then $f$ is called a ring isomorphism from $U$ to $V$.

Example: Let $U={\langle}Z, +, {\times}{\rangle}$ be the ring of integers and $V={\langle}N_m, \mathsf{+}_m, \mathsf{\times}_m{\rangle}$ be the ring of integers modulo $m$, where $N_m=\{0, 1, 2, {\cdots}, m-1\}$. Here $\mathsf{+}_m$ and $\mathsf{\times}_m$ are modular addition and modular multiplication.

That is, for ${\forall}x, y{\in}Z$: $x\mathsf{+}_my=(x+y)\, \mod\, m,\,\,x\mathsf{\times}_my=(x{\times}y)\, \mod\, m$

Let $f:Z{\rightarrow}N_m$ be a function defined by $f(x)=x\, \mod\, m$ for any $x{\in}Z$.

Prove that $f$ is a ring epimorphism from $U$ to $V$.

Proof: For any $x_1, x_2{\in}Z$, we have \[f(x_1+x_2)=(x_1+x_2)\, \mod\, m=x_1\, \mod\, m\mathsf{+}_mx_2\, \mod\, m\]

\[f(x_1{\times}x_2)=(x_1{\times}x_2)\, \mod\, m=x_1\, \mod\, m\mathsf{\times}_mx_2\, \mod\, m\] That is: \[f(x_1+x_2)=f(x_1)\mathsf{+}_mf(x_2),\,\,f(x_1{\times}x_2)=f(x_1)\mathsf{\times}_mf(x_2)\]

Therefore: $f$ is a ring homomorphism from $U$ to $V$.

Further, for ${\forall}n{\in}N_m$, there exists $n{\in}Z$ satisfying $f(n)=n\, \mod\, m$, so $f$ is surjective, and therefore $f$ is a ring epimorphism.

Fields

Definition: An algebraic system $U={\langle}S, +, *{\rangle}$ is called a field if:

${\langle}S, +{\rangle}$ is an Abelian group;
${\langle}S-\{0\}, *{\rangle}$ is an Abelian group;
The operation $*$ satisfies the distributive law over $+$.

Then $U$ is called a field.

The second condition of the definition implies that a field contains at least two elements.

Therefore, a commutative ring with nonzero elements, a unity element, and an inverse for every nonzero element is a field.

Example: If $Z$, $R$, $Q$, and $C$ are the sets of integers, reals, rationals, and complex numbers respectively, with $+$ and ${\times}$ as ordinary addition and multiplication, then: ${\langle}R, +, {\times}{\rangle}$, ${\langle}Q, +, {\times}{\rangle}$, ${\langle}C, +, {\times}{\rangle}$ are all fields.

However, ${\langle}Z, +, {\times}{\rangle}$ is an integral domain but not a field, because ${\langle}Z-\{0\}, {\times}{\rangle}$ is not a group: except for 1 and -1, all other nonzero elements have no inverse.

Theorem: Every field satisfies the cancellation law.

Proof: Let ${\langle}R, +, *{\rangle}$ be a field. For any $a, b, c{\in}R$ with $a{\neq}0$, if $a*b=a*c$, then: $a^{-1}*a*b=a^{-1}*a*c$. Therefore: $b=c$.

Theorem: Every field is an integral domain.

Theorem: Every finite integral domain is a field.