ma/cs 6a2014-15/1term/ma006a/class19.pdf11/12/2014 2 reminder: what we already know about groups...
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Ma/CS 6a Class 19: Isomorphisms and Subgroups
By Adam Sheffer
Reminder: A Group
A group consists of a set πΊ and a binary operation β, satisfying the following.
β¦ Closure. For every π₯, π¦ β πΊ, we have π₯ β π¦ β πΊ.
β¦ Associativity. For every π₯, π¦, π§ β πΊ, we have π₯ β π¦ β π§ = π₯ β π¦ β π§ .
β¦ Identity. The exists π β πΊ, such that for every π₯ β πΊ, we have
π β π₯ = π₯ β π = π₯.
β¦ Inverse. For every π₯ β πΊ there exists π₯β1 β πΊ such that π₯ β π₯β1 = π₯β1 β π₯ = π.
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Reminder: What We Already Know About Groups Given a group with a set πΊ:
β¦ The multiplication table of πΊ is a Latin square.
β¦ The identity is unique.
β¦ For each π β πΊ, there is a unique inverse πβ1.
β¦ For π, π β πΊ, the equation ππ₯ = π has a unique solution.
Reminder: Orders
The order of a group is the number of elements in its set πΊ.
The order of an element π β πΊ is the least positive integer π that satisfies ππ = 1.
Rotation 90β
2 20 1
(under multiplication mod 3)
5
(under integer addition)
4 2 β
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Reminder: A Group of 2 Γ 2 Matrices 2 Γ 2 matrices of the form
πΌ π½0 1
where πΌ β 1,2 and π½ β 0,1,2 .
The operation is matrix multiplication πππ 3.
The group is of order 6: 1 00 1
1 10 1
1 20 1
2 00 1
2 10 1
2 20 1
The Multiplication Table
πΌ =1 00 1
π =1 10 1
π =1 20 1
π =2 00 1
π =2 10 1
π =2 20 1
π° πΉ πΊ πΏ π π
πΌ πΌ π π π π π
π π π πΌ π π π
π π πΌ π π π π
π π π π πΌ π π
π π π π π πΌ π
π π π π π π πΌ
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Symmetries of a Triangle
The six symmetries of the triangle form a group (under composition).
Another Multiplication Table
π π π π π π
π π π π π₯ π¦ π§
π π π π π¦ π§ π₯
π π π π π§ π₯ π¦
π₯ π₯ π§ π¦ π π π
π¦ π¦ π₯ π§ π π π
π§ π§ π¦ π₯ π π π
π π π
π₯ π¦ π§
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π° πΉ πΊ πΏ π π
πΌ πΌ π π π π π
π π π πΌ π π π
π π πΌ π π π π
π π π π πΌ π π
π π π π π πΌ π
π π π π π π πΌ
π π π π π π
π π π π π₯ π¦ π§
π π π π π¦ π§ π₯
π π π π π§ π₯ π¦
π₯ π₯ π§ π¦ π π π
π¦ π¦ π₯ π§ π π π
π§ π§ π¦ π₯ π π π
Isomorphisms
πΊ1, πΊ2 β two groups of the same order.
A bijection π½: πΊ1 β πΊ2 is an isomorphism if for every π, π β πΊ1, we have
π½ ππ = π½ π π½ π .
(i.e., after reordering, we have the same multiplication tables)
When such an isomorphism exists, πΊ1 and πΊ2 are said to be isomorphic, and write πΊ1 β πΊ2.
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Isomorphism Over the Reals
Problem. Prove that the following groups are isomorphic:
β¦ The set of real numbers β under addition.
β¦ The set of positive real numbers β+ under multiplication.
Proof.
β¦ Use the functions ππ₯: β β β+ and log π₯ : β+
β β as bijections between the two sets.
β¦ For π₯, π¦ β β we have ππ₯ππ¦ = ππ₯+π¦.
β¦ For π₯, π¦ β β+, we have log π₯ + log π¦ = log π₯π¦.
Isomorphism Between β€4 and β€5+
Problem. Are the following two groups isomorphic?
β¦ The set β€4 = *0,1,2,3+ under addition πππ 4.
β¦ The set β€5+ = 1,2,3,4 under multiplication
πππ 5.
Solution.
β¦ Yes. Use the following bijection of β€4 β β€5+.
0 β 1 1 β 2 2 β 4 3 β 3
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Cyclic Groups
A group πΊ is cyclic if there exists an element π₯ β πΊ such that every member of πΊ is a power of π₯.
We say that π₯ generates πΊ.
What is the order of π₯? |πΊ|.
An infinite group πΊ is cyclic if there exists an element π₯ β πΊ such that
πΊ = β¦ , π₯β2, π₯β1, 1, π₯, π₯2, π₯3, β¦
Cyclic Groups?
Are the following groups cyclic?
β¦ Integers under addition.
Yes! It is generated by the integer 1 (which is not the identity element).
β¦ The symmetries of the triangle.
No. There are no generators.
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Cyclic Groups?
Are the following groups cyclic?
β¦ The positive reals β+ under multiplication.
No. For example, nothing can generate 1.
β¦ The aforementioned group of elements of the
form πΌ π½0 1
.
No. Because it is isomorphic to the triangle symmetry group.
Finite Cyclic Groups
Any cyclic group of a finite order π with generator π can be written as
1, π, π2, β¦ , ππβ1 .
For integers π and 0 β€ π < π, we have πππ+π = ππ .
Where did we already encounter such a group?
β¦ Integers πππ π under addition.
β¦ The generator is 1 and the group is 0,1,2, β¦ , π β 1 .
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Isomorphic Cyclic Groups
Claim. All of the cyclic groups of a finite order π are isomorphic. We refer to this group as πΆπ.
Proof. Consider two such cyclic groups πΊ1 = 1, π, π2, π3, β¦ , ππβ1 , πΊ2 = 1, β, β2, β3, β¦ , βπβ1 .
β¦ Consider the bijection π½: πΊ1 β πΊ2 satisfying
π½ ππ = βπ.
β¦ This is an isomorphism since π½ ππππ = π½ ππ+π = βπ+π = βπβπ = π½ ππ π½ ππ .
Simple Groups
A trivial group is a group that contains only one element β an identity element.
A simple group is a non-trivial group that does not contain any other βwell-behavedβ subgroups in it.
The finite simple groups are, in a certain sense, the "basic building blocks" of all finite groups.
β¦ Somewhat similar to the way prime numbers are the basic building blocks of the integers.
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Classification of Finite Simple Groups
βOne of the most important mathematical achievements of the 20th century was the collaborative effort, taking up more than 10,000 journal pagesβ (Wikipedia).
Written by about 100 authors!
Theorem. Every finite simple group is isomorphic to one of the following groups:
β¦ A cyclic group.
β¦ An alternating group.
β¦ A simple Lie group.
β¦ One of the 26 sporadic groups.
The Monster Group
One of the 26 sporadic groups is the monster group.
It has an order of 808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000.
The 6 sporadic groups that are not βcontainedβ in the monster group are called the happy family.
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Direct Product
πΊ1, πΊ2 - two groups with identities 11, 12.
The direct product πΊ1 Γ πΊ2 consists of the ordered pairs π, π where π β πΊ1 and π β πΊ2.
The direct product is a group:
β¦ The group operation is π, π π, π = ππ, ππ .
β¦ The identity element is 11, 12 .
β¦ The inverse π, π β1 is πβ1, πβ1 .
β¦ The order of πΊ1 Γ πΊ2 is πΊ1 πΊ2 .
Direct Product Example
Problem. Is πΆ6 isomorphic to πΆ2 Γ πΆ3?
Solution.
β¦ Write πΆ2 = 1, π and πΆ3 = 1, β, β2 .
β¦ Then πΆ2 Γ πΆ3 consists of 1,1 , 1, β , 1, β2 , π, 1 , π, β , (π, β2) .
β¦ πΆ2 Γ πΆ3 is isomorphic to πΆ6 iff it is cyclic.
β¦ It is cyclic, since it is generated by π, β .
π, β 1 = π, β , π, β 2 = 1, β2 , π, β 3 = π, 1 , π, β 4 = 1, β , π, β 5 = π, β2 , π, β 6 = 1,1 ,
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Another Direct Product Example
Problem. Is πΆ8 isomorphic to πΆ2 Γ πΆ4?
Solution.
β¦ Write πΆ2 = 1, π and πΆ4 = 1, β, β2, β3 .
β¦ Then πΆ2 Γ πΆ4 consists of * 1,1 , 1, β , 1, β2 , 1, β3 ,
π, 1 , π, β , π, β2 , π, β3 +.
β¦ πΆ2 Γ πΆ4 is isomorphic to πΆ8 iff it is cyclic.
β¦ It is not. The orders of the elements are 1,4,2,4,2,4,2,4, respectively.
Cyclic Inner Products of Cyclic Groups
Claim. If π, π are relatively prime positive integers, then πΆπ Γ πΆπ β πΆππ.
Proof. Write πΆπ = 1, π, π2, β¦ , ππβ1 and πΆπ = 1, β, β2, β¦ , βπβ1 .
It suffices to prove that π, β generates πΆπ Γ πΆπ.
π, β π = 1,1 if and only if π|π and π|π.
Recall: GCD π, π = 1 implies LCM π, π= ππ.
Thus, the order of π, β is ππ.
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Cyclic Inner Products of Cyclic Groups Claim. If π, π are relatively prime positive
integers, then πΆπ Γ πΆπ β πΆππ.
Proof. Write πΆπ = 1, π, π2, β¦ , ππβ1 and πΆπ = 1, β, β2, β¦ , βπβ1 .
We proved that π, β is of order ππ.
It remains to show that for every 0 β€ π < π < ππ, we have π, β π β π, β π.
If π, β π = π, β π, multiplying both sides by π, β βπ implies π, β πβπ = 1,1 .
Contradiction to the order of π, β ! So π, β generates ππ distinct elements.
Subgroups
A subgroup of a group πΊ is a group with the same operation as πΊ, and whose set of members is a subset of πΊ.
Find a subgroup of the group of integers under addition.
β¦ The subset of even integers.
β¦ The subset β¦ , β2π, βπ, 0, π, 2π, . . for any integer π > 1.
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Subgroups of a Symmetry Group
Problem. Find a subgroup of the symmetries of the square.
No action Rotation 90β Rotation 180β Rotation 270β
Vertical flip Horizontal flip Diagonal flip Diagonal flip 2
Subgroups of a Symmetry Group
Problem. Find a subgroup of the subgroup.
No action Rotation 90β Rotation 180β Rotation 270β
Vertical flip Horizontal flip Diagonal flip Diagonal flip 2
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Subgroups of a Symmetry Group
No action Rotation 90β Rotation 180β Rotation 270β
Vertical flip Horizontal flip Diagonal flip Diagonal flip 2
Subgroup Conditions
Problem. Let πΊ be a group, and let π» be a non-empty subset of πΊ such that
β¦ C1. If π₯, π¦ β π» then π₯π¦ β π».
β¦ ππ. If π₯ β π» then π₯β1 β π».
Prove that π» is a subgroup.
β¦ Closure. By C1.
β¦ Inverse. By C2.
β¦ Associativity. By the associativity of πΊ.
β¦ Identity. By C2, π₯, π₯β1 β π». By C1, we have 1 = π₯π₯β1 β π».
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Finite Subgroup Conditions
Problem. Let πΊ be a finite group, and let π» be a non-empty subset of πΊ such that
β¦ C1. If π₯, π¦ β π» then π₯π¦ β π».
β¦ ππ. If π₯ β π» then π₯β1 β π».
Prove that π» is a subgroup.
Proof. Consider π₯ β π».
Since πΊ is finite, the series 1, π₯, π₯2, π₯3, β¦ has two identical elements π₯π = π₯π with π < π.
Multiply both side by π₯βπβ1 (in πΊ) to obtain π₯β1 = π₯πβπβ1 = π₯π₯π₯ β― π₯ β π».
The End: A Noahβs Ark Joke The Flood has receded and the ark is safely aground atop Mount Ararat; Noah tells all the animals to go forth and multiply. Soon the land is teeming with every kind of living creature in abundance, except for snakes. Noah wonders why. One morning two miserable snakes knock on the door of the ark with a complaint. βYou havenβt cut down any trees.β Noah is puzzled, but does as they wish. Within a month, you canβt walk a step without treading on baby snakes. With difficulty, he tracks down the two parents. βWhat was all that with the trees?β βAh,β says one of the snakes, βyou didnβt notice which species we are.β Noah still looks blank. βWeβre adders, and we can only multiply using logs.β
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