Group Theory

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

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

Group Axioms

1) Closure. For any x,yGx, y \in G , xyx*y is also in GG.

2) Associativity. For any x,y,zGx, y, z \in G , we have (xy)z=x(yz) (x *y)*z = x*(y*z) .

3) Identity. There exists an eG e \in G , such that ex=xe e * x = x * e for any xGx \in G . We say that ee is an identity element of GG.

4) Inverse. For any xGx \in G, there exists a yGy \in G such that xy=e=yxxy = e = yx . We say that yy is an inverse of xx.

Although the associativity condition only specifies three elements, it can be generalized to arbitrarily many elements. For example, to show that (w(xy))z=w(x(yz)) (w*(x*y))*z=w*(x*(y*z)), we apply the condition twice to obtain: (w(xy))z=w((xy)z)=w(x(yz)). (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 nn elements of GG: a1a2.ana_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 a1a2ana_1a_2\ldots a_n. However, the order of the elements matters, since it is generally not true that xy=yxxy = yx for all x,yGx,y \in G . In fact, we say that the group G is abelian if for any x,yGx, y \in G, xy=yxxy = 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) Q \mathbb{Q} , the set of rational numbers, with the group operation of addition.
2) R× \mathbb{R}^\times , the set of non-zero real numbers, with the group operation of multiplication.
3) Sn S_n, the set of bijective functions [n][n] [n] \rightarrow [n] , where [n]={1,2,,n} [n] = \{1, 2, \ldots, n \} , with the group operation of function composition.
4) Zn \mathbb{Z}_n, the set of integers {0,1,,n1} \{0, 1, \ldots, n-1\} , with group operation of addition modulo nn.
5) Zn× \mathbb{Z}_n ^\times , the set of integers {1an1:gcd(a,n)=1} \{ 1 \leq a \leq n-1: \gcd(a,n)=1 \} , with group operation of multiplication modulo nn.

Let xGx\in G be an element, with an inverse yy . For any mZm \in \mathbb{Z}, we define:

xm={xxx(m terms)if m>0,e,if m=0,yyy(m terms)if m<0 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 gGg \in G and m,nZm, n \in \mathbb{Z} , then gm+n=gmgn g^{m+n} = g^m g^n and (gm)n=gmn \left( g^m \right)^n =g^{mn} .

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

Worked examples

1. Let GG be a group. Then the identity element eG e \in G is unique. Also, every element xG x \in G has a unique inverse, which we shall denote by x1 x^{-1} .

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

Similarly, let yy and yy' be inverses of xx. Then

y=ye=y(xy)=(yx)y=ey=y. 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 (x1)1=x.(x^{-1})^{-1} = x. Furthermore, we can show that (xm)1=xm (x^m)^{-1} = x^{-m} .


2. If x,yGx, y \in G have inverses x1,y1 x^{-1}, y^{-1} respectively, what is the inverse of xy xy?

Solution: The inverse of the product xyx * y is given by y1x1y^{-1} * x^{-1}. Indeed, we have (xy)(y1x1)=x(yy1)x1=xex1=e(x * y)*(y^{-1}*x^{-1})=x(y*y^{-1})x^{-1} =xex^{-1} =e and likewise, (y1x1)(xy)=e(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 (xx), and then you put on your shoes (y) (y) . At then end of the day, you have to take off your shoes (y1) (y^{-1} ) , and then take off your socks (x1) ( x^{-1}) . Hence (xy)1=y1x1 (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 GG be a group. If g,h,hGg, h, h' \in G and gh=ghgh = gh', then h=hh=h'. Likewise, if g,g,hG g, g', h \in G and gh=ghgh = g'h, then g=gg = g'.

Solution: For the first statement, the equation gh=ghgh = gh' gives: g1(gh)=g1(gh)g^{-1}(gh) = g^{-1}(gh'), so (g1g)h=(g1g)h(g^{-1}g)h = (g^{-1}g)h' and thus h=hh = h'. For the second statement, multiply h1h^{-1} on the right.


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


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!n! elements.
4) There are nn elements.
5) There are ϕ(n)\phi(n) elements.

Note by Calvin Lin
5 years, 4 months ago

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