Groups - mathSight

It is useful, when reading the below, to have a background knowledge of Set Theory.

A set, by itself, is lacking in structure. To enable a structure that permits, say, solutions to equations or defines symmetries we need additional properties to be bestowed on the set. One such property is that of forming a Group.

A Group is an algebraic structure that has a single law of composition on a set. A law of composition takes two elements from the set and maps these to a third element.

Definition: Let ${S}$ is a set then a law of composition ${*}$ is defined as:

$\displaystyle *:S\; \times \; S \rightarrow S \ \ \ \ \ (1)$

By reference to set theory the is a map of ${(x,y)\in A\times A}$ to ${z \in A}$ .

To be a Group, the law of composition ${*}$ must, for all elements ${x}$ and ${y}$ , map these to an element ${z}$ in the same Group. If it does this, then ${*}$ is said to be closed.

There are a few other requirements for a set with a law of composition to form a Group.

Definition: Identity element ${e}$ . Let ${G}$ be a set and ${*}$ a law of composition.

$\displaystyle \exists e \in G \; \forall x\in G \; [x*e = x \; \land \; e*x = x] \ \ \ \ \ (2)$

There exists an identity element ${e}$ within the set ${G}$ that when composed with any element ${x}$ of ${G}$ maps to ${x}$ .

Definition: Inverse element ${x^{-1}}$ . Let ${G}$ be a set and ${*}$ a law of composition.

$\displaystyle \forall x \in G \; \exists x^{-1} \in G \; [x*x^{-1}=e \; \land \; x^{-1}*x = e] \ \ \ \ \ (3)$

There exists an inverse element in the set for each element that when composed with that element produces the identity element.

Definition: The composition law is associative. Let ${G}$ be a set.

$\displaystyle \forall x \in G, \forall y \in G, \forall z \in G \; [x*(y*z)=(x*y)*z] \ \ \ \ \ (4)$

The order of application of the composite law does not change the mapping.

If we have a set with a law of composition that is closed and for which an identity element exists, each element has an inverse and the law is associative then we have a Group.

The identity element ${e}$ is unique in a Group.

Proof:
Let ${e_1\in G}$ and ${e_2\in G}$ . Let ${*}$ be a law of composition on the Group ${G}$ .
By the definition of the identity element:
${e_1=e_1*e_2}$ where ${e_2}$ is the right identity but
${e_1*e_2=e_2}$ where ${e_1}$ is the left identity ${\implies e_1 = e_2}$ $\Box$

The inverse of an element ${x}$ is unique.

Proof:
Let ${x_1^{-1}=x_1^{-1}e\implies x_1^{-1}=x_1^{-1}(x x_2^{-1})}$
${\implies x_1^{-1}=(x_1^{-1}x)x_2^{-1}\implies x_1^{-1}=ex_2^{-1}}$
${\implies x_1^{-1}=x_2^{-1}}$ $\Box$

If the composition $*$ is commutative on the set $G$ , that is $\forall x,y \in G [x * y = y*x]$ then the Group is said be abelian.

A common Group is that of the set of integers $\mathbb{Z}$ under addition $+$ . In this Group, the identity element $e=0$ . That is $x+0=x$ . The inverse element for $x=-x$ . This is verified by $x+(-x)=0$ which is indeed defined as the identity element.

The set $\mathbb{Z}$ is not a Group under the law of multiplication $.$ . Whilst there is an identity element $1$ such that $x.1=x$ there is no inverse in the Group as for example, $1/x$ is not an element of $\mathbb{Z}$ .

Further, if we have the set of all real numbers $\mathbb{R}$ then this is not a Group under multiplication. This is because $0$ is an element of the real numbers and there is no inverse element for $0$ . If we exclude $0$ from the set $\mathbb{R}$ then the $\mathbb{R}$ is a group under multiplication with an identity element of $1$ and an inverse for $x$ of $1/x$ .

Since Group structures allow solutions to equations, the lack of an inverse element under multiplication to the real numbers explains why the ‘high school’ bad practice of solving equations leads to a contradiction. $0 \times 3 = 0 \times 4$ , cancel the $0$ and we are left with $3=4$ . This is a contradiction because we do not ‘cancel’ anything. The correct way to solve this equation is to apply the inverse of $0$ to both sides so that this $0$ and its inverse equals the identity element. The problem is, the inverse of $0$ is not defined and so the equation cannot be solved. Put simply, you cannot divide through by $0$ !.