A **group** is a set $G$ together with a function $G\times G \rightarrow G$, which is denoted by $(x,y) \mapsto x * y$ satisfying the following properties (also known as the group axioms).

## Group Axioms

1)

Closure. For any $x, y \in G$, $x*y$ is also in $G$.2)

Associativity. For any $x, y, z \in G$, we have $(x *y)*z = x*(y*z)$.3)

Identity. There exists an $e \in G$, such that $e * x = x * e$ for any $x \in G$. We say that $e$ is an identity element of $G$.4)

Inverse. For any $x \in G$, there exists a $y \in G$ such that $xy = e = yx$. We say that $y$ is an inverse of $x$.

Although the associativity condition only specifies three elements, it can be generalized to arbitrarily many elements. For example, to show that $(w*(x*y))*z=w*(x*(y*z))$, we apply the condition twice to obtain: $(w * (x * y)) * z = w * ((x * y) * z) = w * (x * (y * z)).$

Thus, we can drop the parentheses altogether and speak of the product of $n$ elements of $G$: $a_1 * a_2 * .\ldots * a_n$, since it does not matter how we arrange the parentheses. Sometimes, we even drop the $*$ altogether and write it as $a_1a_2\ldots a_n$. However, the order of the elements matters, since it is generally not true that $xy = yx$ for all $x,y \in G$. In fact, we say that the group G is **abelian** if for any $x, y \in G$, $xy = yx$.

To specify a group, we have to state what the set is, along with the group operation. The following are common examples of some groups that you may have seen before.

1) $\mathbb{Q}$, the set of rational numbers, with the group operation of addition.

2) $\mathbb{R}^\times$, the set of non-zero real numbers, with the group operation of multiplication.

3) $S_n$, the set of bijective functions $[n] \rightarrow [n]$, where $[n] = \{1, 2, \ldots, n \}$, with the group operation of function composition.

4) $\mathbb{Z}_n$, the set of integers $\{0, 1, \ldots, n-1\}$, with group operation of addition modulo $n$.

5) $\mathbb{Z}_n ^\times$, the set of integers $\{ 1 \leq a \leq n-1: \gcd(a,n)=1 \}$, with group operation of multiplication modulo $n$.

Let $x\in G$ be an element, with an inverse $y$. For any $m \in \mathbb{Z}$, we define:

$x^m = \begin{cases} x*x*\ldots x \quad (m \mbox{ terms}) & \mbox{if } m > 0, \\ e, & \mbox{if } m = 0, \\ y * y * \ldots * y \quad (-m \mbox{ terms}) & \mbox{if } m < 0 \\ \end{cases}$

It is routine, but rather tedious, to show that the exponential laws of integers similarly hold.

For any $g \in G$ and $m, n \in \mathbb{Z}$, then $g^{m+n} = g^m g^n$ and $\left( g^m \right)^n =g^{mn}$.

The **order** of a group $G$, is the number of elements in $G$, which we denote by $\lvert G \rvert$ like in set notation.

## 1. Let $G$ be a group. Then the identity element $e \in G$ is unique. Also, every element $x \in G$ has a unique inverse, which we shall denote by $x^{-1}$.

Solution: Let $e$ and $e'$ be identities. Then by definition, we get: $e' = e * e' = e$.

Similarly, let $y$ and $y'$ be inverses of $x$. Then

$y=y*e=y*(x*y')=(y*x)*y' =e*y' =y'.$

Note that the inverse of the inverse of x is precisely x itself. In symbolic form, we get $(x^{-1})^{-1} = x.$ Furthermore, we can show that $(x^m)^{-1} = x^{-m}$.

## 2. If $x, y \in G$ have inverses $x^{-1}, y^{-1}$ respectively, what is the inverse of $xy$?

Solution: The inverse of the product $x * y$ is given by $y^{-1} * x^{-1}$. Indeed, we have $(x * y)*(y^{-1}*x^{-1})=x(y*y^{-1})x^{-1} =xex^{-1} =e$ and likewise, $(y^{-1}*x^{-1})*(x*y)=e$.

A simple way to remember this property, is to think about how you wear your socks and shoes. You first put on your socks ($x$), and then you put on your shoes $(y)$. At then end of the day, you have to take off your shoes $(y^{-1} )$, and then take off your socks $( x^{-1})$. Hence $(xy)^{-1} = y^{-1} x^{-1}$. Trying to take off your socks while your shoes are on, is going to be very difficult.

## 3. [Cancellation Law]. Let $G$ be a group. If $g, h, h' \in G$ and $gh = gh'$, then $h=h'$. Likewise, if $g, g', h \in G$ and $gh = g'h$, then $g = g'$.

Solution: For the first statement, the equation $gh = gh'$ gives: $g^{-1}(gh) = g^{-1}(gh')$, so $(g^{-1}g)h = (g^{-1}g)h'$ and thus $h = h'$. For the second statement, multiply $h^{-1}$ on the right.

## 4. What is the order of each of the 5 groups listed above?

Solution:

1) There are infinitely many elements. (In fact, there are countably many elements.)

2) There are infinitely many elements. (In fact, there are uncountably many elements.)

3) There are $n!$ elements.

4) There are $n$ elements.

5) There are $\phi(n)$ elements.

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