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# This note has been used to help create the Group Theory wiki

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.

## Worked examples

### 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.

Note by Calvin Lin
3 years, 6 months ago

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