Groups

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

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 $a=e$. $\blacksquare$

Subgroups and Group Extensions

Definition: Let $(G,\cdot )$ be a group. If $S\subseteq G$ and $S$ forms a group under the same operation $\cdot$ then $(S,\cdot )$ is said to be a Subgroup of $(G,\cdot )$. If $S$ is a subgroup of $G$ then we write $S\leq G$.

Definition: Let $(G,\cdot )$ be a group. If $(H,\cdot )$ is a group such that $G\subseteq H$ then $(H,\cdot )$ is said to be a Group Extension of $(G,\cdot )$.

For example, consider the group of complex numbers under the operation of standard addition, $(\mathbb {C} ,+)$. We know that the set of real numbers is a subset of the set of complex numbers, that is, $\mathbb {R} \subset \mathbb {C}$ and so the group of real numbers under the operation of standard addition, $(\mathbb {R} ,+)$ is a subgroup of $(\mathbb {C} ,+)$ and $(\mathbb {C} ,+)$ is a group extension of $(\mathbb {R} ,+)$.

We will now look at a nice theorem which tells us that to determine if $(S,\cdot )$ is a subgroup of $(G,\cdot )$ where $S\subseteq G$, that then we only need to check two of the four group axioms for verification

Theorem 1: If $(G,\cdot )$ is a group with the identity $e\in G$ and $S\subseteq G$ then $(S,\cdot )$ is a subgroup of $(G,\cdot )$ if and only if $S$ is closed under $\cdot$ and for all $a\in S$ there exists an $a^{-1}\in S$ such that $a\cdot a^{-1}=e$ and $a^{-1}\cdot a=e$.

• Proof: Let $(G,\cdot )$ be a group with the identity $e\in G$ of $\cdot$ and let $S\subseteq G$.

• $\Rightarrow$ Suppose that $(S,\cdot )$ is a subgroup of $(G,\cdot )$. Then by definition, $(S,\cdot )$ is a group itself and satisfies all of the group axioms - namely that $S$ is closed under the operation $\cdot$ and that for all $a\in S$ there exists an $a^{-1}\in S$ such that $a\cdot a^{-1}=e$ and $a^{-1}\cdot a=e$.

• $\Leftarrow$ Now suppose that $S$ is closed under $\cdot$ and that for all $a\in S$ there exists an $a^{-1}\in S$ such that $a\cdot a^{-1}=e$ and $a^{-1}\cdot a=e$. These are precisely two of the group axioms we have looked at, and to show that $(S,\cdot )$ is a subgroup of $(G,\cdot )$ we only need to show that the other two axioms hold.

• First suppose that $a,b,c\in S$ and that $a\cdot (b\cdot c)\neq (a\cdot b)\cdot c$, that is, suppose that $\cdot$ is not associative on $S$. Since $S\subseteq G$ we must have that $a\cdot (b\cdot c)\neq (a\cdot b)\cdot c$ for this particular $a,b,c\in G$ which contradicts the associativity of $\cdot$ on the group $G$. Hence $\cdot$ must actually be associative on $G$.

• Now since $S$ is closed under $\cdot$ and for $a\in S$ there exists an $a^{-1}\in S$ such that $a\cdot a^{-1}=e$ and $a^{-1}\cdot a=e$ we must have that $e\in S$ and furthermore, $a\cdot e=a$ and $e\cdot a=a$.

• Therefore $(S,\cdot )$ is a group, and in particular since $S\subseteq G$ we have that $(S,\cdot )$ is a subgroup of $(G,\cdot )$. $\blacksquare$

Theorem 2: If $(G,\cdot )$ is a group with the identity $e\in G$ of $\cdot$ and $S\subseteq G$ then $(S,\cdot )$ is a subgroup of $(G,\cdot )$ if and only if $H\neq \emptyset$ and for all $a,b\in H$ we have that $a\cdot b^{-1}\in H$.

• Proof: $\Rightarrow$ If $(S,\cdot )$ is a subgroup of $(G,\cdot )$ then this direction is trivial.

• $\Leftarrow$ Suppose that $S\neq \emptyset$ and for all $a,b\in S$ we have that $a\cdot b^{-1}\in S$. Since $S\neq \emptyset$ there exists an $a\in S$. So $a\cdot a^{-1}=e\in S$.

• Now if $a\in S$, then since $e\in S$ we have that $e\cdot a^{-1}=a^{-1}\in S$. So if $a\in S$ then $a^{-1}\in H$.

• Lastly, if $a,b\in S$ then $a,b^{-1}\in S$. Thus $a\cdot (b^{-1})^{-1}=a\cdot b\in S$. So $S$ is closed under the operation $\cdot$. Thus $(S,\cdot )$ is a subgroup of $(G,\cdot )$. $\blacksquare$

Order of an element in a group

Definition: Let $(G,*)$ be a group and let $a\in G$. The Order of $a$ denoted by $|a|$ or ${\text{ord}}(a)$ is the smallest positive integer $m$ such that $a^{m}=\underbrace {a\cdot a\cdot ...\cdot a} _{m:{\text{many factors}}}=e$ (where $e$ is the identity element of $G$). If no such $m$ exists then $a$ is said to have order $\infty$.

If the operation is multiplicative in nature then we usually define the order of $a\in G$ as above. If the operation is instead additive in nature then we define the order of $a\in G$ as the smallest positive integer $m$ such that $ma=\underbrace {(a+a+...+a)} _{m\;{\text{many}}}=e$ or $\infty$ if no such positive integer $m$ exists.

Example 1

If $G$ is any group with identity $e$ then the order of $e$ is $1$.

Example 2

Consider the group $(\mathbb {Z} _{5},+)$ where $+$ is defined for all $x,y\in \mathbb {Z} _{5}=\{0,1,2,3,4\}$ to be:

${\begin{aligned}\quad x+y=(x+y)\mod 5\end{aligned}}$

The order of $e=0$ is trivially $1$. The order of $1$ is $5$ since:

${\begin{aligned}\quad 1^{5}=(1+1+1+1+1)=0\mod 5\end{aligned}}$

The order of $2$ is also $5$. In fact, the orders of $3$ and $4$ are also $5$.

Example 3

Consider the group $(\mathbb {Z} _{6},+)$ where $+$ is defined for all $x,y\in \mathbb {Z} _{6}=\{0,1,2,3,4,5\}$ to be:

${\begin{aligned}\quad x+y=(x+y)\mod 6\end{aligned}}$

You should verify that the order of $0$ is $1$, the order of $1$ is $6$, the order of $2$ is $3$, the order of $3$ is $2$, the order of $4$ is $3$, and the order of $5$ is $6$

Example 4

Consider the group $(\mathbb {R} ,+)$. Then every nonzero $x\in \mathbb {R}$ has order infinity since the equation $mx=0$, $m\geq 1$, has no solution in $\mathbb {R} \setminus \{0\}$.

Example 5

Consider the group $(\mathbb {R} \setminus \{0\},\cdot )$. The order of $-1$ is $2$ since $(-1)^{2}=1$.

Basic Theorems Regarding the Order of Elements in a Group

Proposition 1: Let $(G,\cdot )$ be a group. Then:

a) $\mathrm {order} (a)=1$ if and only if $a=e$ where $e$ is the identity element of $G$.
b) If $a\in G$ then $\mathrm {ord} (a^{-1})=\mathrm {ord} (a)$.

c) If $a,b\in G$ then $\mathrm {ord} (a\cdot b)=\mathrm {ord} (b\cdot a)$. _

• Proof of a) $\Rightarrow$ Suppose that $\mathrm {ord} (a)$. Then $a^{1}=e$. So $a=e$.

• $\Leftarrow$ The smallest positive integer such that $e^{n}=e$ is $n=1$. So $\mathrm {ord} (e)=1$. $\blacksquare$

• Proof of b) Let $a\in G$. Suppose that $a$ has finite order, say $\mathrm {ord} (a)=n$. Then $n$ is the smallest positive integer such that $a^{n}=e$. So $a^{-n}=e$. So $\mathrm {ord} (a^{-1})\leq n$. If $\mathrm {ord} (a^{-1})=m<n$ then $a^{-m}=e$ implies that $a^{m}=e$, and since $m<n=\mathrm {ord} (a)$ we have arrived at a contradiction. Thus $\mathrm {ord} (a^{-1})=\mathrm {ord} (a)$.

• Now suppose that $a$ has infinite order. If $a^{-1}$ has finite order, say $\mathrm {ord} (a^{-1})=m<\infty$ then $a^{-m}=e$ implies that $a^{m}=e$ - contradicting $a$ having infinite order. Thus $a^{-1}$ must also have infinite order.

• Proof of c) Let $a,b\in G$. First, suppose that $\mathrm {ord} (a\cdot b)=n<\infty$. Then $n$ is the smallest positive integer such that $(a\cdot b)^{n}=e$. So $(a\cdot b)^{n}\cdot a=a$. Now observe that:

${\begin{aligned}\quad (a\cdot b)^{n}\cdot a=\underbrace {(a\cdot b)\cdot (a\cdot b)\cdot ...\cdot (a\cdot b)} _{n\;{\text{factors}}}\cdot a=a\cdot \underbrace {(b\cdot a)\cdot (b\cdot a)\cdot ...\cdot (b\cdot a)} _{n\;{\text{factors}}}=a\cdot (b\cdot a)^{n}\end{aligned}}$

• Therefore $(a\cdot b)^{n}\cdot a=a\cdot (b\cdot a)^{n}$. But $(a\cdot b)^{n}=e$, so $a=a\cdot (b\cdot a)^{n}$. Thus $(b\cdot a)^{n}=e$. So $\mathrm {ord} (b\cdot a)\leq n=\mathrm {ord} (a\cdot b)$.

• By symmetry, we see that $\mathrm {ord} (a\cdot b)\leq \mathrm {ord} (b\cdot a)$. Thus $\mathrm {ord} (a\cdot b)=\mathrm {ord} (b\cdot a)$.

• Now suppose that $\mathrm {ord} (a\cdot b)$ is infinite. If $\mathrm {ord} (b\cdot a)=n<\infty$ then $(b\cdot a)^{n}=e$. By the same argument above, we see that $(a\cdot b)^{n}=e$ - contradicting $a\cdot b$ having infinite order. Thus $\mathrm {ord} (b\cdot a)$ is infinite. $\blacksquare$

Licensing

Content obtained and/or adapted from:

• Groups, mathonline.wikidot.com under a CC BY-SA license
• Basic Theorems Regarding Groups, mathonline.wikidot.com under a CC BY-SA license
• Subgroups and Group Extensions, mathonline.wikidot.com under a CC BY-SA license