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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