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1

Some Finite Groups

Chapter 3

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One element GroupLet G be an one element group. Then its only one element must be the identity element e , so

G = { e }Note. Because multiplication ee = e , the inverse of e is e .That means we always have

e−1 = eand

1 0{1},

0 1G G

� �� �= = � �� �

� �

are one element group examples

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3

Two elements GroupLet G be a two element group. Because that one elementmust be the identity element e , we can assume that

G = { e , a}Question: What is multiplication result of aa ? If aa = a then

(aa)a−1 = aa−1

From associative lawa(aa−1) = aa−1

And ae = e a = e Because that a and e are different, this is a contradiction.Therefore we must have

aa = e or a2 = e

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Multiplication TableLet G = { e , a} be a two element group. Then we have the following 4 multiplication results:

ee = e ea = aae = a aa = e

We can write all those results in a table as follows:

e a

e

a(L)

(R)

e a

a e

(L) means left side elements, (R) means right side elements. In this multiplication table, every row and every column has no duplicated elements

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5

Two elements Group ExamplesIn real number case G = { 1, −1 } is a two element group. In 2×2 matrices, let

be a two elements group.From above argument, we know that

1 0,

0 1a b

c dG

� �� �� � � �� �� � � �� �� � � �� �

=

1 0*

0 1

1 00 1

a b a b

c d c d

aa bc ab bdca dc cb dd

� � � � � �=� � � � � �

� � � � � �

� � � �+ +=� � � �

+ +� � � �

6

2

2

So we get equation system

1001

One possible solution is 0, 0, 1, 1. So

The other solutions second matrices

1 0 1 0,

0 1 0 1

1 0 1 0 0 1, ,

0 1 0 1

are

G

a bc

ab db

ac dc

d bc

b c a d

� �� � � �= � �� � � �−� � � �� �

− −� � � �� �

� + =� + =��

+ =�� + =

�

�

= = = =

�

−

�−� � � 1 0� �� �� �

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Three elements GroupLet G = { e, a, b} be a three element group. We try to obtain its multiplication table

e a b

e

a

b

(L)

(R)

e a b

a ? ?

b ? ?

We first consider what is ab=? It is either e or b.

If ab=b then must a=e and this is impossible.

So we must have ab=e. Similarly ba=e.

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Now we obtain its multiplication table

e a b

e

a

b

(L)

(R)

e a b

a b e

b e b

From this table, we have

ab=e, ba=e

a2 = b , b2 = a ,

And a3= aa2 = ab=e, b3= bb2 = ba=e,

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Three elements Group Examples{ }

( )( )( )( )( )( )

1 3 1 32 2 2 2

1 3 1 32 2 2 2

21 3 1 32 2 2 2

21 3 1 32 2 2 2

31 32 2

31 32 2

Group is a three elements group.

We can verify the multiplication table as follw

1, ,

1

1

oing:

1

G i i

i i

i i

i i

i

i

+ −

+ −

+ −

− − +

+

− −

= − −

− − =

− = −

= −

− =

=

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Another 3 elements Group exampleLet Z = {All integers}, define equivalent relation ~ in Z as following: For any two integers m and n if 3|(m−n),we say that m ~ nThen we have three equivalent subclasses.

a = { 3k : k∈Z }b = { 3k+1 : k∈Z }c = { 3k+2 : k∈Z }

Let set G = { a, b, c} Now we define operation + in G: Pick any m ∈ a and any n ∈ b , we can see m+n ∈ b .So we define a + b = b and b + a = b Pick any m ∈ a and any n ∈ c , we can see m+n ∈ c .So we define a + c = c and c + a = c

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Pick any m ∈ a and any n ∈ a, we can see m+n ∈ a .So we define a + a = aPick any m ∈ b and any n ∈ b , we can see m+n ∈ c .So we define b + b = cPick any m ∈ c and any n ∈ c, we can see m+n ∈ b .So we define c + c = bPick any m ∈ b and any n ∈ c , we can see m+n ∈ a .So we define b + c = a and c + b = aThe addition table is

a b c

a

b

c

(L)

(R)

a b c

b c a

c a b

{G, +} is group and

identity element is a

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Order of elementLet G be a group and a∈G. Let n be the smallest positive integer such that an = e, then we say that a has order n.Theorem: If a has order n, then set G= { e, a, a2 , a3…, an−1 }is a group. Such a is called a generator of G. We say that G is generated by a. We can see that G commutesTheorem: Any one element group is generated by identity. Any two elements group is generated by an element of order 2.Any three elements group is generated by an element of order 3

G = {1, −1} is generated by −1 (−1 has order 2)G = {1, i, −1, −i} is generated by i or −i (both have order 4)

{ }1 3 1 3 1 3 1 3

2 2 2 2 2 2 2 2is gene1, , rated by or G i i i i− + − − − + − −=

(both have order 3)