chapter 11:fundamental theorem of finite abelian groups
TRANSCRIPT
Abstract Algebra (I)
Chapter 11: Fundamental Theorem ofFinite Abelian Groups
The Fundamental Theorem
The Isomorphism Classes of Abelian Groups
Dr. Ahmed El-MabhouhFirst Semester 2019-2020
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 1 / 1
Theorem (Fundamental Theorem of Finite Abelian Groups)
Every finite Abelian group is a direct product of cyclic groups ofprime-power order. Moreover, the number of terms in the productand the orders of the cyclic groups are uniquely determined by thegroup. That is:
G ≈ Zp1n1 ⊕ Zp2n2 ...⊕ Zpknk
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 2 / 1
The Isomorphism Classes of Abelian Groups
Let’s look at groups whose orders have the form pk , where p is primeand k ≤ 4. In general, there is one group of order pk for each set ofpositive integers whose sum is k .
Definition
(Partition)A set of positive integers n1, n2, ..., nt is called a partition of k ifk = n1 + n2 + ... + nt .
In this caseZpn1 ⊕ Zpn2 ...⊕ Zpnt
is an Abelian group of order pk .
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 3 / 1
Partition
Figure: Isomorphism Classes
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 4 / 1
Now, we know how to construct all the Abelian groups of primepower order, pk .
we move to the problem of constructing all Abelian groups of acertain order n, where n has two or more distinct prime divisors.
We begin by writing n in prime-power decomposition formn = pn11 pn22 ...pnkk .
Next, we individually form all Abelian groups of orderpn11 , then pn22 ,and so on.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 5 / 1
Example
Find all non isomorphic Abelian groups of order n = 1176.
Note that 1176 = 23.3.72.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 6 / 1
Example
Find all non isomorphic Abelian groups of order n = 1176.
Note that 1176 = 23.3.72.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 6 / 1
Example
Let G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64}under multiplication modulo 65. Write G as an external product ofcyclic groups.
Proof.
Note that G = 16 = 24.
Partition of 4 is: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.
So we have 5 classesZ16
Z8 ⊕ Z2
Z4 ⊕ Z4
Z4 ⊕ Z2 ⊕ Z2
Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2
G is isomorphic to only one of these 5 groups, which one?
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 7 / 1
Example
Let G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64}under multiplication modulo 65. Write G as an external product ofcyclic groups.
Proof.
Note that G = 16 = 24.
Partition of 4 is: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.
So we have 5 classesZ16
Z8 ⊕ Z2
Z4 ⊕ Z4
Z4 ⊕ Z2 ⊕ Z2
Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2
G is isomorphic to only one of these 5 groups, which one?
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 7 / 1
Example
Let G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64}under multiplication modulo 65. Write G as an external product ofcyclic groups.
Proof.
Note that G = 16 = 24.
Partition of 4 is: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.
So we have 5 classesZ16
Z8 ⊕ Z2
Z4 ⊕ Z4
Z4 ⊕ Z2 ⊕ Z2
Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2
G is isomorphic to only one of these 5 groups, which one?
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 7 / 1
Example
Let G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64}under multiplication modulo 65. Write G as an external product ofcyclic groups.
Proof.
Note that G = 16 = 24.
Partition of 4 is: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.
So we have 5 classesZ16
Z8 ⊕ Z2
Z4 ⊕ Z4
Z4 ⊕ Z2 ⊕ Z2
Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2
G is isomorphic to only one of these 5 groups, which one?
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 7 / 1
Example
Let G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64}under multiplication modulo 65. Write G as an external product ofcyclic groups.
Proof.
Note that G = 16 = 24.
Partition of 4 is: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.
So we have 5 classesZ16
Z8 ⊕ Z2
Z4 ⊕ Z4
Z4 ⊕ Z2 ⊕ Z2
Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2
G is isomorphic to only one of these 5 groups, which one?
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 7 / 1
Example
Let G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64}under multiplication modulo 65. Write G as an external product ofcyclic groups.
Proof.
Note that G = 16 = 24.
Partition of 4 is: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.
So we have 5 classesZ16
Z8 ⊕ Z2
Z4 ⊕ Z4
Z4 ⊕ Z2 ⊕ Z2
Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2
G is isomorphic to only one of these 5 groups, which one?
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 7 / 1
Table of orders of elements of G
Proof.
To decide which one, we calculate the orders of the elements ofG .
From the table of orders, we rule out Z16, Z8 ⊕ Z2 andZ2 ⊕ Z2 ⊕ Z2 ⊕ Z2. Why?
It remains Z4 ⊕ Z4 and Z4 ⊕ Z2 ⊕ Z2
We rule out Z4 ⊕ Z2 ⊕ Z2 since it has more that 3 elements oforder 2 but G has only 3 elements or order 3.
Therefore G is isomorphic to Z4 ⊕ Z4.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 8 / 1
Table of orders of elements of G
Proof.
To decide which one, we calculate the orders of the elements ofG .
From the table of orders, we rule out Z16, Z8 ⊕ Z2 andZ2 ⊕ Z2 ⊕ Z2 ⊕ Z2. Why?
It remains Z4 ⊕ Z4 and Z4 ⊕ Z2 ⊕ Z2
We rule out Z4 ⊕ Z2 ⊕ Z2 since it has more that 3 elements oforder 2 but G has only 3 elements or order 3.
Therefore G is isomorphic to Z4 ⊕ Z4.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 8 / 1
Table of orders of elements of G
Proof.
To decide which one, we calculate the orders of the elements ofG .
From the table of orders, we rule out Z16, Z8 ⊕ Z2 andZ2 ⊕ Z2 ⊕ Z2 ⊕ Z2. Why?
It remains Z4 ⊕ Z4 and Z4 ⊕ Z2 ⊕ Z2
We rule out Z4 ⊕ Z2 ⊕ Z2 since it has more that 3 elements oforder 2 but G has only 3 elements or order 3.
Therefore G is isomorphic to Z4 ⊕ Z4.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 8 / 1
Table of orders of elements of G
Proof.
To decide which one, we calculate the orders of the elements ofG .
From the table of orders, we rule out Z16, Z8 ⊕ Z2 andZ2 ⊕ Z2 ⊕ Z2 ⊕ Z2. Why?
It remains Z4 ⊕ Z4 and Z4 ⊕ Z2 ⊕ Z2
We rule out Z4 ⊕ Z2 ⊕ Z2 since it has more that 3 elements oforder 2 but G has only 3 elements or order 3.
Therefore G is isomorphic to Z4 ⊕ Z4.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 8 / 1
Table of orders of elements of G
Proof.
To decide which one, we calculate the orders of the elements ofG .
From the table of orders, we rule out Z16, Z8 ⊕ Z2 andZ2 ⊕ Z2 ⊕ Z2 ⊕ Z2. Why?
It remains Z4 ⊕ Z4 and Z4 ⊕ Z2 ⊕ Z2
We rule out Z4 ⊕ Z2 ⊕ Z2 since it has more that 3 elements oforder 2 but G has only 3 elements or order 3.
Therefore G is isomorphic to Z4 ⊕ Z4.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 8 / 1
Table of orders of elements of G
Proof.
To decide which one, we calculate the orders of the elements ofG .
From the table of orders, we rule out Z16, Z8 ⊕ Z2 andZ2 ⊕ Z2 ⊕ Z2 ⊕ Z2. Why?
It remains Z4 ⊕ Z4 and Z4 ⊕ Z2 ⊕ Z2
We rule out Z4 ⊕ Z2 ⊕ Z2 since it has more that 3 elements oforder 2 but G has only 3 elements or order 3.
Therefore G is isomorphic to Z4 ⊕ Z4.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 8 / 1
Example
Let G = {1, 8, 17, 19, 26, 28, 37, 44, 46, 53, 62, 64, 71, 73,82, 89, 91, 98, 107, 109, 116, 118, 127, 134}under multiplication modulo 135. Since G has order 24, it isisomorphic to one of the following:
Z8 ⊕ Z3∼= Z24,
Z4 ⊕ Z2 ⊕ Z3∼= Z12 ⊕ Z3,
Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3∼= Z6 ⊕ Z2 ⊕ Z2,
decide which one.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 9 / 1
Proof.
Since |8| = 12 we rule out the third one,Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3
∼= Z6 ⊕ Z2 ⊕ Z2,
Since |109| = |134| = 2, G has at least 2 elements of order 2and Z8 ⊕ Z3
∼= Z24, has only one element of order 2, Why? werule out the first one.
Therefore G ∼=Z4 ⊕ Z2 ⊕ Z3∼= Z12 ⊕ Z3.
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 10 / 1
Corollary
If m divides the order of a finite Abelian group G , then G has asubgroup of order m.
Example
Let G be an Abelian group of order 72, we show that G always hasa subroup of order 12.
Proof.
Since |G | = 72 = 23.32, then G is isomorphic to one of thefollowing:
Therefore, we show each of these groups has a subgroup of order12
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 11 / 1
Exercises of Ch 11
1-1012151924-26
A. EL-Mabhouh (Islamic University of Gaza Faculty of Science Department of Mathematics )Abstract Algebra II Second Semester 2018-2019 12 / 1