Abstract Algebra: Groups

Recall that an operation $\cdot$ on $S$ is said to be associative if for all $a,b,c\in S$ we have that $a\cdot (b\cdot c)=(a\cdot b)\cdot c$ and $\cdot$ is said to be commutative if for all $a,b\in S$ we have that $a\cdot b=b\cdot a$.

An element $e\in S$ is the identity element of $S$ under $\cdot$ if for all $a\in S$ we have that $a\cdot e=a$ and $e\cdot a=a$.

We can now begin to describe our first type of algebraic structures known as groups, which are a set $G$ equipped with a binary operation $\cdot$ that is associative, contains an identity element, and contains inverse elements under $\cdot$ for each element in $G$.

Definition: A Group is a pair $(G,\cdot )$ where $G$ is a set and $\cdot$ is a binary operation on $G$ with the following properties:

1. For all $a,b,c\in G$, $a\cdot (b\cdot c)=(a\cdot b)\cdot c$ (Associativity of $\cdot$).
2. There exists an $e\in G$ such that for all $a\in G$, $a\cdot e=a$ and $e\cdot a=a$ (The existence of an Identity Element).
3. For all $a\in G$ there exists an $a^{-1}\in G$ such that $a\cdot a^{-1}=e$ and $a^{-1}\cdot a=e$ (The existence of inverses).

Furthermore, if $G$ is a finite set then the group $(G,\cdot )$ is said to be a Finite Group and if $G$ is an infinite set then the group $(G,\cdot )$ is said to be an Infinite Group. More generally, the Order of $(G,\cdot )$ (or **Size of $(G,\cdot )$) is the size of $G$ and is denoted $|G|$.

When we use the multiplication symbol $\cdot$ to denote the operation on $G$, we often call $G$ a “multiplicative group”. When the operation of the group is instead denoted by $+$ (instead of $\cdot$) then we often call $G$ an “additive group”, and we write the inverse of each $a\in G$ as $-a$ (instead of $a^{-1}$).

Some of the sets and binary operations we have already seen can be considered groups. For example, $(\mathbb {R} ,+)$ is a group under standard addition $+$ since the sum of any two real numbers is a real number, $+$, is associative, an additive identity $0\in \mathbb {R}$ exists and inverse elements exist for every $a\in \mathbb {R}$ (namely $-a\in \mathbb {R}$).

Furthermore, $(\mathbb {Z} ,+)$ is also a group under the operation of standard addition since the sum of any two integers is an integer, addition is associative, the additivity identity is $0\in \mathbb {Z}$, and for all $a\in \mathbb {Z}$ we have $-a\in \mathbb {Z}$ as additive inverses.

We will examine many other (more interesting) groups later on, but for now, let's look at an example of a set and a binary operation that does NOT form a group.

Example 1

Consider the set of integers $\mathbb {Z}$ and define $*$ for all $a,b\in \mathbb {Z}$ by:

${\begin{aligned}\quad a*b=a+2b\end{aligned}}$

(Where the $+$ on the righthand side is usual addition of numbers). We will show that $(\mathbb {Z} ,*)$ is NOT a group by showing that $*$ is not associative. Let $a,b,c\in \mathbb {Z}$. Then $*$ is not associative since:

${\begin{aligned}\quad a*(b*c)=a*(b+2c)=a+2(b+2c)=a+2b+4c\end{aligned}}$

${\begin{aligned}\quad (a*b)*c=(a+2b)*c=(a+2b)+2c=a+2b+2c\end{aligned}}$

Clearly $a*(b*c)\neq (a*b)*c$ so $\mathbb {Z}$ does not form a group under the operation $*$.

Basic Theorems Regarding Groups

Recall that a group $(G,\cdot )$ is a set $G$ with a binary operation $\cdot$ such that:

• 1) $\cdot$ is associative, i.e., for all $a,b,c\in G$, $a\cdot (b\cdot c)=(a\cdot b)\cdot c)$.

• 2) There exists an identity element $e\in G$ such that $a\cdot e=a=e\cdot a$ for all $a\in G$.

• 3) For each $a\in G$ there exists an $a^{-1}\in G$ such that $a\cdot a^{-1}=a^{-1}\cdot a=e$.

We will now look at some rather basic results regarding groups which we can derive from the group axioms above.

Proposition 1: Let $(G,\cdot )$ be a group and let $e$ be the identity for this group. Then:

a) The identity element $e\in G$ is unique.
b) For each $a\in G$, the corresponding inverse $a^{-1}\in G$ is unique.
c) For each $a\in G$, $(a^{-1})^{-1}=a$.
d) For all $a,b\in G$, $(a\cdot b)^{-1}=b^{-1}\cdot a^{-1}$.
e) For all $a,b\in G$, if $a\cdot b=e$ then $a=b^{-1}$ and $b=a^{-1}$.

f) If $a^{2}=a$ then $a=e$.

• Proof of a) Suppose that $e$ and $e'$ are both identities for $\cdot$. Then:

${\begin{aligned}\quad e=e\cdot e=e\cdot e'=e'\cdot e'=e'\end{aligned}}$

• Therefore $e=e'$ so the identity for $\cdot$ is unique. $\blacksquare$

• Proof of b) Suppose that $a^{-1}\in G$ and $a^{-1'}\in G$ are both inverses for $a\in G$ under $\cdot$. Then:

${\begin{aligned}\quad a^{-1}=a^{-1}\cdot e=a^{-1}\cdot (a\cdot a^{-1'})=(a^{-1}\cdot a)*a^{-1}=e\cdot a^{-1'}=a^{-1'}\end{aligned}}$

• Therefore $a^{-1}=a^{-1'}$ so the inverse for $a$ is unique. $\blacksquare$

• Proof of c) Let $a\in G$. Then $(a^{-1})^{-1}$ is the inverse to $a^{-1}$. However, the inverse to $a^{-1}$ is $a$ and by (b) we have shown that the inverse of each element in $G$ is unique. Therefore $a=(a^{-1})^{-1}$. $\blacksquare$

• Proof of d) If we apply the operation $\cdot$ between $b^{-1}\cdot a^{-1}$ and $(a\cdot b)$ we get:

${\begin{aligned}\quad (a\cdot b)\cdot [b^{-1}\cdot a^{-1}]&=a\cdot [(b\cdot b^{-1})\cdot a^{-1}]\\\quad &=a\cdot [e\cdot a^{-1}]\\\quad &=a\cdot a^{-1}\\\quad &=e\end{aligned}}$

• Therefore the inverse of $(a\cdot b)$ is $b^{-1}\cdot a^{-1}$. We also have that the invere of $(a\cdot b)$ is $(a\cdot b)^{-1}$. By (b), the inverse of $(a\cdot b)$ is unique and so:

${\begin{aligned}\quad (a\cdot b)^{-1}=b^{-1}\cdot a^{-1}\quad \blacksquare \end{aligned}}$

• Proof of e) Suppose that $a\cdot b=e$. Then:

${\begin{aligned}\quad a\cdot b&=e\\\quad (a\cdot b)\cdot b^{-1}&=e\cdot b^{-1}\\\quad a\cdot (b\cdot b^{-1})&=b^{-1}\\\quad a\cdot e&=b^{-1}\\\quad a&=b^{-1}\end{aligned}}$

• Similarly:

${\begin{aligned}\quad a\cdot b&=e\\\quad a^{-1}\cdot (a\cdot b)&=a^{-1}\cdot e\\\quad (a^{-1}\cdot a)\cdot b&=a^{-1}\\\quad e\cdot b&=a^{-1}\\\quad b&=a^{-1}\quad \blacksquare \end{aligned}}$

• Proof of f) Suppose that $a^{2}=a\cdot a=a$. Then:

${\begin{aligned}\quad a^{2}&=a\\\quad a\cdot a&=a\\\quad a^{-1}\cdot (a\cdot a)&=a^{-1}\cdot a\\\quad (a^{-1}\cdot a)\cdot a&=e\\\quad e\cdot a&=e\\\quad a&=e\end{aligned}}$

• Hence $a=e$. Alternatively we see that if $a\cdot a=a$ then the inverse of $a$ with respect to $\cdot$ is $e$, that is $a^{-1}=e$. Multiplying both sides of this equation by $a$ gives us that $a=e$. $\blacksquare$

Cancellation Law

We will now look at another important property of groups called the cancellation law.

Theorem 1 (The Cancellation Law for Groups): Let $(G,\cdot )$ be a group and let $a,b,c\in G$. If $a\cdot b=a\cdot c$ or $b\cdot a=c\cdot a$ then $b=c$.

• Proof: Let $a^{-1}\in G$ denote the inverse of $a$ under $\cdot$. Suppose that $a\cdot b=a\cdot c$. Then:

${\begin{aligned}\quad a\cdot b&=a\cdot c\\\quad (a^{-1}\cdot a)\cdot b&=(a^{-1}\cdot a)\cdot c\\\quad e\cdot b&=e\cdot c\\\quad b&=c\end{aligned}}$

• Similarly, suppose now that $b\cdot a=c\cdot a$. Then:

${\begin{aligned}\quad b\cdot a&=c\cdot a\\\quad (b\cdot a)\cdot a^{-1}&=(c\cdot a)\cdot a^{-1}\\\quad b\cdot (a\cdot a^{-1})&=c\cdot (a\cdot a^{-1})\\\quad b\cdot e&=c\cdot e\\\quad b&=c\quad \blacksquare \end{aligned}}$

It is very important to note that the cancellation law holds with regards to the operation $\cdot$ for any group $(G,\cdot )$. We will see that the cancellation law does not necessarily hold with respect to an operation on a set when we look at algebraic structures with two defined operations.

It is also important to note that if $a\cdot b=c\cdot a$ or $b\cdot a=a\cdot c$ then we cannot necessarily deduce that $b=c$ because we would then require the additional property that $\cdot$ is commutative which is not one of the group axioms (but instead one of the Abelian group axioms).

Licensing

Content obtained and/or adapted from:

• Abstract Algebra, mathonline.wikidot.com under a CC BY-SA license