Anatomy of the mulitplicative group Greg Martin University of - - PowerPoint PPT Presentation

anatomy of the mulitplicative group
SMART_READER_LITE
LIVE PREVIEW

Anatomy of the mulitplicative group Greg Martin University of - - PowerPoint PPT Presentation

Questions about Z Other invariants of Z Distribution of ( n ) Elements and subgroups n n Anatomy of the mulitplicative group Greg Martin University of British Columbia Pacific Northwest Number Theory Conference University of Idaho


slide-1
SLIDE 1

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Anatomy of the mulitplicative group

Greg Martin

University of British Columbia Pacific Northwest Number Theory Conference University of Idaho May 19, 2012

slides can be found on my web page www.math.ubc.ca/∼gerg/index.shtml?slides

Anatomy of the mulitplicative group Greg Martin

slide-2
SLIDE 2

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Outline

1

What do we want to know about multiplicative groups (mod n)?

2

Distribution of the number of prime factors of n

3

Other invariants of the multiplicative groups

4

Counting certain elements, and subgroups, of multiplicative groups

Anatomy of the mulitplicative group Greg Martin

slide-3
SLIDE 3

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The main characters

Notation

The quotient ring Z/nZ will be denoted by Zn. It enjoys both addition and multiplication.

If we ignore multiplication:

The additive group Z+

n is the set Zn with the ring’s addition

  • peration. It is always a cyclic group with n elements.

If we instead ignore addition:

The multiplicative group Z×

n is the set (Zn)× of invertible

elements in Zn, with the ring’s multiplication operation. It is a finite abelian group with φ(n) elements.

Anatomy of the mulitplicative group Greg Martin

slide-4
SLIDE 4

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The main characters

One ring to rule them all . . .

The quotient ring Z/nZ will be denoted by Zn. It enjoys both addition and multiplication.

If we ignore multiplication:

The additive group Z+

n is the set Zn with the ring’s addition

  • peration. It is always a cyclic group with n elements.

If we instead ignore addition:

The multiplicative group Z×

n is the set (Zn)× of invertible

elements in Zn, with the ring’s multiplication operation. It is a finite abelian group with φ(n) elements.

Anatomy of the mulitplicative group Greg Martin

slide-5
SLIDE 5

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The main characters

One ring to rule them all . . .

The quotient ring Z/nZ will be denoted by Zn. It enjoys both addition and multiplication.

If we ignore multiplication:

The additive group Z+

n is the set Zn with the ring’s addition

  • peration. It is always a cyclic group with n elements.

If we instead ignore addition:

The multiplicative group Z×

n is the set (Zn)× of invertible

elements in Zn, with the ring’s multiplication operation. It is a finite abelian group with φ(n) elements.

Anatomy of the mulitplicative group Greg Martin

slide-6
SLIDE 6

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The main characters

One ring to rule them all . . .

The quotient ring Z/nZ will be denoted by Zn. It enjoys both addition and multiplication.

If we ignore multiplication:

The additive group Z+

n is the set Zn with the ring’s addition

  • peration. It is always a cyclic group with n elements.

If we instead ignore addition:

The multiplicative group Z×

n is the set (Zn)× of invertible

elements in Zn, with the ring’s multiplication operation. It is a finite abelian group with φ(n) elements.

Anatomy of the mulitplicative group Greg Martin

slide-7
SLIDE 7

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

How to find the structure of Z×

n

Chinese remainder theorem, and primitive roots

If the prime factorization of n is pr1

1 × · · · × prk k , then

n ∼

= Z×

pr1

1 × · · · × Z×

p

rk k

∼ = Z+

pr1−1

1

(p1−1) × · · · × Z+ p

rk−1 k

(pk−1).

Confession

I’m lying slightly about the prime p = 2. I’ll keep doing so throughout the talk when making general statements about Z×

n .

Example (with n = 11!)

11! ∼

= Z×

28 ⊕ Z× 34 ⊕ Z× 52 ⊕ Z× 7 ⊕ Z× 11

∼ = (Z+

2 ⊕ Z+ 64) ⊕ Z+ 54 ⊕ Z+ 20 ⊕ Z+ 6 ⊕ Z+ 10

Anatomy of the mulitplicative group Greg Martin

slide-8
SLIDE 8

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

How to find the structure of Z×

n

Chinese remainder theorem, and primitive roots

If the prime factorization of n is pr1

1 × · · · × prk k , then

n ∼

= Z×

pr1

1 × · · · × Z×

p

rk k

∼ = Z+

pr1−1

1

(p1−1) × · · · × Z+ p

rk−1 k

(pk−1).

Confession

I’m lying slightly about the prime p = 2. I’ll keep doing so throughout the talk when making general statements about Z×

n .

Example (with n = 11!)

11! ∼

= Z×

28 ⊕ Z× 34 ⊕ Z× 52 ⊕ Z× 7 ⊕ Z× 11

∼ = (Z+

2 ⊕ Z+ 64) ⊕ Z+ 54 ⊕ Z+ 20 ⊕ Z+ 6 ⊕ Z+ 10

Anatomy of the mulitplicative group Greg Martin

slide-9
SLIDE 9

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

How to find the structure of Z×

n

Chinese remainder theorem, and primitive roots

If the prime factorization of n is pr1

1 × · · · × prk k , then

n ∼

= Z×

pr1

1 × · · · × Z×

p

rk k

∼ = Z+

pr1−1

1

(p1−1) × · · · × Z+ p

rk−1 k

(pk−1).

Confession

I’m lying slightly about the prime p = 2. I’ll keep doing so throughout the talk when making general statements about Z×

n .

Example (with n = 11!)

11! ∼

= Z×

28 ⊕ Z× 34 ⊕ Z× 52 ⊕ Z× 7 ⊕ Z× 11

∼ = (Z+

2 ⊕ Z+ 64) ⊕ Z+ 54 ⊕ Z+ 20 ⊕ Z+ 6 ⊕ Z+ 10

Anatomy of the mulitplicative group Greg Martin

slide-10
SLIDE 10

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

How to find the structure of Z×

n

Chinese remainder theorem, and primitive roots

If the prime factorization of n is pr1

1 × · · · × prk k , then

n ∼

= Z×

pr1

1 × · · · × Z×

p

rk k

∼ = Z+

pr1−1

1

(p1−1) × · · · × Z+ p

rk−1 k

(pk−1).

Confession

I’m lying slightly about the prime p = 2. I’ll keep doing so throughout the talk when making general statements about Z×

n .

Example (with n = 11!)

11! ∼

= Z×

28 ⊕ Z× 34 ⊕ Z× 52 ⊕ Z× 7 ⊕ Z× 11

∼ = (Z+

2 ⊕ Z+ 64) ⊕ Z+ 54 ⊕ Z+ 20 ⊕ Z+ 6 ⊕ Z+ 10

Anatomy of the mulitplicative group Greg Martin

slide-11
SLIDE 11

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

How to find the structure of Z×

n

Chinese remainder theorem, and primitive roots

If the prime factorization of n is pr1

1 × · · · × prk k , then

n ∼

= Z×

pr1

1 × · · · × Z×

p

rk k

∼ = Z+

pr1−1

1

(p1−1) × · · · × Z+ p

rk−1 k

(pk−1).

Confession

I’m lying slightly about the prime p = 2. I’ll keep doing so throughout the talk when making general statements about Z×

n .

Example (with n = 11!)

11! ∼

= Z×

28 ⊕ Z× 34 ⊕ Z× 52 ⊕ Z× 7 ⊕ Z× 11

∼ = (Z+

2 ⊕ Z+ 64) ⊕ Z+ 54 ⊕ Z+ 20 ⊕ Z+ 6 ⊕ Z+ 10

Anatomy of the mulitplicative group Greg Martin

slide-12
SLIDE 12

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

How to find the structure of Z×

n

Chinese remainder theorem, and primitive roots

If the prime factorization of n is pr1

1 × · · · × prk k , then

n ∼

= Z×

pr1

1 × · · · × Z×

p

rk k

∼ = Z+

pr1−1

1

(p1−1) × · · · × Z+ p

rk−1 k

(pk−1).

Confession

I’m lying slightly about the prime p = 2. I’ll keep doing so throughout the talk when making general statements about Z×

n .

Example (with n = 11!)

11! ∼

= Z×

28 ⊕ Z× 34 ⊕ Z× 52 ⊕ Z× 7 ⊕ Z× 11

∼ = (Z+

2 ⊕ Z+ 64) ⊕ Z+ 54 ⊕ Z+ 20 ⊕ Z+ 6 ⊕ Z+ 10

Anatomy of the mulitplicative group Greg Martin

slide-13
SLIDE 13

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Two standard forms

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm.

Example (with n = 11!)

11! ∼

= Z+

2 ⊕ Z+ 2 ⊕ Z+ 2 ⊕ Z+ 2 ⊕ Z+ 60 ⊕ Z+ 8640

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ where each qj is

  • prime. (unique up to reordering)

Anatomy of the mulitplicative group Greg Martin

slide-14
SLIDE 14

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Two standard forms

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm.

Example (with n = 11!)

11! ∼

= Z+

2 ⊕ Z+ 2 ⊕ Z+ 2 ⊕ Z+ 2 ⊕ Z+ 60 ⊕ Z+ 8640

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ where each qj is

  • prime. (unique up to reordering)

Anatomy of the mulitplicative group Greg Martin

slide-15
SLIDE 15

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Two standard forms

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm.

Example (with n = 11!)

11! ∼

= Z+

2 ⊕ Z+ 2 ⊕ Z+ 2 ⊕ Z+ 2 ⊕ Z+ 60 ⊕ Z+ 8640

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ where each qj is

  • prime. (unique up to reordering)

Anatomy of the mulitplicative group Greg Martin

slide-16
SLIDE 16

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Two standard forms

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm.

Example (with n = 11!)

11! ∼

= (Z+

2 )4 ⊕ Z+ 60 ⊕ Z+ 8640

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ where each qj is

  • prime. (unique up to reordering)

Anatomy of the mulitplicative group Greg Martin

slide-17
SLIDE 17

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Two standard forms

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm.

Example (with n = 11!)

11! ∼

= (Z+

2 )4 ⊕ Z+ 60 ⊕ Z+ 8640

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ where each qj is

  • prime. (unique up to reordering)

Anatomy of the mulitplicative group Greg Martin

slide-18
SLIDE 18

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Two standard forms

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm.

Example (with n = 11!)

11! ∼

= (Z+

2 )4 ⊕ Z+ 4 ⊕ Z+ 64 ⊕ Z+ 3 ⊕ Z+ 27 ⊕ (Z+ 5 )2

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ where each qj is

  • prime. (unique up to reordering)

Anatomy of the mulitplicative group Greg Martin

slide-19
SLIDE 19

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Reasons to look at the multiplicative group

We study Z×

n because of its:

Ubiquity

Modular arithmetic shows up everywhere in number theory.

Typicality

n is representative of finite abelian groups in general.

Fun exercise

Every finite abelian group is a subgroup of Z×

n for infinitely

many integers n.

Anatomy of the mulitplicative group Greg Martin

slide-20
SLIDE 20

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Reasons to look at the multiplicative group

We study Z×

n because of its:

Ubiquity

Modular arithmetic shows up everywhere in number theory.

Typicality

n is representative of finite abelian groups in general.

Fun exercise

Every finite abelian group is a subgroup of Z×

n for infinitely

many integers n.

Anatomy of the mulitplicative group Greg Martin

slide-21
SLIDE 21

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Reasons to look at the multiplicative group

We study Z×

n because of its:

Ubiquity

Modular arithmetic shows up everywhere in number theory.

Typicality

n is representative of finite abelian groups in general.

Fun exercise

Every finite abelian group is a subgroup of Z×

n for infinitely

many integers n.

Anatomy of the mulitplicative group Greg Martin

slide-22
SLIDE 22

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Reasons to look at the multiplicative group

We study Z×

n because of its:

Ubiquity

Modular arithmetic shows up everywhere in number theory.

Typicality

n is representative of finite abelian groups in general.

Fun exercise

Every finite abelian group is a subgroup of Z×

n for infinitely

many integers n.

Anatomy of the mulitplicative group Greg Martin

slide-23
SLIDE 23

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Questions to ask about Z×

n

If we “choose n at random”, what is the distribution of: the number of invariant factors? the largest invariant factor? the number of terms in the primary decomposition? the largest term in the primary decomposition? the number of elements of order 2 (square roots of 1 (mod n))? (and generalizations) the number of subgroups?

Choosing n at random means:

Choose n uniformly at random from an initial interval {1, 2, . . . , x}, understand the distribution as a function of x, and see what happens as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-24
SLIDE 24

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Questions to ask about Z×

n

If we “choose n at random”, what is the distribution of: the number of invariant factors? the largest invariant factor? the number of terms in the primary decomposition? the largest term in the primary decomposition? the number of elements of order 2 (square roots of 1 (mod n))? (and generalizations) the number of subgroups?

Choosing n at random means:

Choose n uniformly at random from an initial interval {1, 2, . . . , x}, understand the distribution as a function of x, and see what happens as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-25
SLIDE 25

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Questions to ask about Z×

n

If we “choose n at random”, what is the distribution of: the number of invariant factors? the largest invariant factor? the number of terms in the primary decomposition? the largest term in the primary decomposition? the number of elements of order 2 (square roots of 1 (mod n))? (and generalizations) the number of subgroups?

Choosing n at random means:

Choose n uniformly at random from an initial interval {1, 2, . . . , x}, understand the distribution as a function of x, and see what happens as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-26
SLIDE 26

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Questions to ask about Z×

n

If we “choose n at random”, what is the distribution of: the number of invariant factors? the largest invariant factor? the number of terms in the primary decomposition? the largest term in the primary decomposition? the number of elements of order 2 (square roots of 1 (mod n))? (and generalizations) the number of subgroups?

Choosing n at random means:

Choose n uniformly at random from an initial interval {1, 2, . . . , x}, understand the distribution as a function of x, and see what happens as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-27
SLIDE 27

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Questions to ask about Z×

n

If we “choose n at random”, what is the distribution of: the number of invariant factors? the largest invariant factor? the number of terms in the primary decomposition? the largest term in the primary decomposition? the number of elements of order 2 (square roots of 1 (mod n))? (and generalizations) the number of subgroups?

Choosing n at random means:

Choose n uniformly at random from an initial interval {1, 2, . . . , x}, understand the distribution as a function of x, and see what happens as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-28
SLIDE 28

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Questions to ask about Z×

n

If we “choose n at random”, what is the distribution of: the number of invariant factors? the largest invariant factor? the number of terms in the primary decomposition? the largest term in the primary decomposition? the number of elements of order 2 (square roots of 1 (mod n))? (and generalizations) the number of subgroups?

Choosing n at random means:

Choose n uniformly at random from an initial interval {1, 2, . . . , x}, understand the distribution as a function of x, and see what happens as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-29
SLIDE 29

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Questions to ask about Z×

n

If we “choose n at random”, what is the distribution of: the number of invariant factors? the largest invariant factor? the number of terms in the primary decomposition? the largest term in the primary decomposition? the number of elements of order 2 (square roots of 1 (mod n))? (and generalizations) the number of subgroups?

Choosing n at random means:

Choose n uniformly at random from an initial interval {1, 2, . . . , x}, understand the distribution as a function of x, and see what happens as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-30
SLIDE 30

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Questions to ask about Z×

n

If we “choose n at random”, what is the distribution of: the number of invariant factors? the largest invariant factor? the number of terms in the primary decomposition? the largest term in the primary decomposition? the number of elements of order 2 (square roots of 1 (mod n))? (and generalizations) the number of subgroups?

Choosing n at random means:

Choose n uniformly at random from an initial interval {1, 2, . . . , x}, understand the distribution as a function of x, and see what happens as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-31
SLIDE 31

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of invariant factors

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The number of invariant factors equals ω(n), the number of distinct prime factors of n. (This is again a slight lie: if n is even, then it might be ω(n) ± 1.)

Theorem (Average order of ω(n))

1 x

  • n≤x

ω(n) = 1

x

  • n≤x
  • p|n

1 = 1

x

  • p≤x
  • n≤x

p|n

1 ∼ 1

x

  • p≤x

x p ∼ log log x.

new improvements in error term (M.–Naslund, 2012+)

Anatomy of the mulitplicative group Greg Martin

slide-32
SLIDE 32

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of invariant factors

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The number of invariant factors equals ω(n), the number of distinct prime factors of n. (This is again a slight lie: if n is even, then it might be ω(n) ± 1.)

Theorem (Average order of ω(n))

1 x

  • n≤x

ω(n) = 1

x

  • n≤x
  • p|n

1 = 1

x

  • p≤x
  • n≤x

p|n

1 ∼ 1

x

  • p≤x

x p ∼ log log x.

new improvements in error term (M.–Naslund, 2012+)

Anatomy of the mulitplicative group Greg Martin

slide-33
SLIDE 33

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of invariant factors

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The number of invariant factors equals ω(n), the number of distinct prime factors of n. (This is again a slight lie: if n is even, then it might be ω(n) ± 1.)

Theorem (Average order of ω(n))

1 x

  • n≤x

ω(n) = 1

x

  • n≤x
  • p|n

1 = 1

x

  • p≤x
  • n≤x

p|n

1 ∼ 1

x

  • p≤x

x p ∼ log log x.

new improvements in error term (M.–Naslund, 2012+)

Anatomy of the mulitplicative group Greg Martin

slide-34
SLIDE 34

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of invariant factors

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The number of invariant factors equals ω(n), the number of distinct prime factors of n. (This is again a slight lie: if n is even, then it might be ω(n) ± 1.)

Theorem (Average order of ω(n))

1 x

  • n≤x

ω(n) = 1

x

  • n≤x
  • p|n

1 = 1

x

  • p≤x
  • n≤x

p|n

1 ∼ 1

x

  • p≤x

x p ∼ log log x.

new improvements in error term (M.–Naslund, 2012+)

Anatomy of the mulitplicative group Greg Martin

slide-35
SLIDE 35

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of invariant factors

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The number of invariant factors equals ω(n), the number of distinct prime factors of n. (This is again a slight lie: if n is even, then it might be ω(n) ± 1.)

Theorem (Average order of ω(n))

1 x

  • n≤x

ω(n) = 1

x

  • n≤x
  • p|n

1 = 1

x

  • p≤x
  • n≤x

p|n

1 ∼ 1

x

  • p≤x

x p ∼ log log x.

new improvements in error term (M.–Naslund, 2012+)

Anatomy of the mulitplicative group Greg Martin

slide-36
SLIDE 36

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of invariant factors

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The number of invariant factors equals ω(n), the number of distinct prime factors of n. (This is again a slight lie: if n is even, then it might be ω(n) ± 1.)

Theorem (Average order of ω(n))

1 x

  • n≤x

ω(n) = 1

x

  • n≤x
  • p|n

1 = 1

x

  • p≤x
  • n≤x

p|n

1 ∼ 1

x

  • p≤x

x p ∼ log log x.

new improvements in error term (M.–Naslund, 2012+)

Anatomy of the mulitplicative group Greg Martin

slide-37
SLIDE 37

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The variance of ω(n)

ω(n) = number of distinct prime factors of n

Theorem (Turán, 1934)

1 x

  • n≤x
  • ω(n) − log log x

2 ∼ log log x. So there can’t be too many integers n with ω(n) far from log log n. For example, the number of integers n ≤ x with

  • ω(n) − log log x
  • > (log log x)0.52 is less than x/(log log x)0.03.

Theorem (Hardy–Ramanujan, 1917)

The normal order of ω(n) is log log n. In other words, if n ∈ {1, 2, . . . , x} is chosen uniformly at random, then the probability that ω(n) ∼ log log n tends to 1 as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-38
SLIDE 38

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The variance of ω(n)

ω(n) = number of distinct prime factors of n

Theorem (Turán, 1934)

1 x

  • n≤x
  • ω(n) − log log x

2 ∼ log log x. So there can’t be too many integers n with ω(n) far from log log n. For example, the number of integers n ≤ x with

  • ω(n) − log log x
  • > (log log x)0.52 is less than x/(log log x)0.03.

Theorem (Hardy–Ramanujan, 1917)

The normal order of ω(n) is log log n. In other words, if n ∈ {1, 2, . . . , x} is chosen uniformly at random, then the probability that ω(n) ∼ log log n tends to 1 as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-39
SLIDE 39

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The variance of ω(n)

ω(n) = number of distinct prime factors of n

Theorem (Turán, 1934)

1 x

  • n≤x
  • ω(n) − log log x

2 ∼ log log x. So there can’t be too many integers n with ω(n) far from log log n. For example, the number of integers n ≤ x with

  • ω(n) − log log x
  • > (log log x)0.52 is less than x/(log log x)0.03.

Theorem (Hardy–Ramanujan, 1917)

The normal order of ω(n) is log log n. In other words, if n ∈ {1, 2, . . . , x} is chosen uniformly at random, then the probability that ω(n) ∼ log log n tends to 1 as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-40
SLIDE 40

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The variance of ω(n)

ω(n) = number of distinct prime factors of n

Theorem (Turán, 1934)

1 x

  • n≤x
  • ω(n) − log log x

2 ∼ log log x. So there can’t be too many integers n with ω(n) far from log log n. For example, the number of integers n ≤ x with

  • ω(n) − log log x
  • > (log log x)0.52 is less than x/(log log x)0.03.

Theorem (Hardy–Ramanujan, 1917)

The normal order of ω(n) is log log n. In other words, if n ∈ {1, 2, . . . , x} is chosen uniformly at random, then the probability that ω(n) ∼ log log n tends to 1 as x → ∞.

Anatomy of the mulitplicative group Greg Martin

slide-41
SLIDE 41

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

A Gaussian distribution?!

Definition (Standard normal distribution/bell curve)

Φ(α) = 1 √ 2π α

−∞

e−t2/2 dt

Theorem (Erd˝

  • s–Kac, 1940)

lim

x→∞

1 x#

  • n ≤ x: ω(n) − log log n

√log log n < α

  • = Φ(α).

“The number of prime factors of n has a normal distribution with mean log log n and standard deviation √log log n.”

Anatomy of the mulitplicative group Greg Martin

slide-42
SLIDE 42

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

A Gaussian distribution?!

Definition (Standard normal distribution/bell curve)

Φ(α) = 1 √ 2π α

−∞

e−t2/2 dt

Theorem (Erd˝

  • s–Kac, 1940)

lim

x→∞

1 x#

  • n ≤ x: ω(n) − log log n

√log log n < α

  • = Φ(α).

“The number of prime factors of n has a normal distribution with mean log log n and standard deviation √log log n.”

Anatomy of the mulitplicative group Greg Martin

slide-43
SLIDE 43

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

A Gaussian distribution?!

Definition (Standard normal distribution/bell curve)

Φ(α) = 1 √ 2π α

−∞

e−t2/2 dt

Theorem (Erd˝

  • s–Kac, 1940)

lim

x→∞

1 x#

  • n ≤ x: ω(n) − log log n

√log log n < α

  • = Φ(α).

“The number of prime factors of n has a normal distribution with mean log log n and standard deviation √log log n.”

Anatomy of the mulitplicative group Greg Martin

slide-44
SLIDE 44

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

A Gaussian distribution?!

Definition (Standard normal distribution/bell curve)

Φ(α) = 1 √ 2π α

−∞

e−t2/2 dt

Theorem (Erd˝

  • s–Kac, 1940)

lim

x→∞

1 x#

  • n ≤ x: ω(n) − log log n

√log log n < α

  • = Φ(α).

“The number of prime factors of n has a normal distribution with mean log log n and standard deviation √log log n.”

Good luck testing this emperically . . .

Even if we could reliably factor numbers around 10100, the quantity log log 10100 isn’t even up to 5.5 yet.

Anatomy of the mulitplicative group Greg Martin

slide-45
SLIDE 45

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

An application: the Erd˝

  • s multiplication table problem

Question:

How many distinct integers are in the N × N multiplication table? × 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 12 14 16 18 20 3 3 6 9 12 15 18 21 24 27 30 4 4 8 12 16 20 24 28 32 36 40 5 5 10 15 20 25 30 35 40 45 50 6 6 12 18 24 30 36 42 48 54 60 7 7 14 21 28 35 42 49 56 63 70 8 8 16 24 32 40 48 56 64 72 80 9 9 18 27 36 45 54 63 72 81 90 10 10 20 30 40 50 60 70 80 90 100

Anatomy of the mulitplicative group Greg Martin

slide-46
SLIDE 46

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

An application: the Erd˝

  • s multiplication table problem

Question:

How many distinct integers are in the N × N multiplication table? × 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 12 14 16 18 20 3 3 6 9 12 15 18 21 24 27 30 4 4 8 12 16 20 24 28 32 36 40 5 5 10 15 20 25 30 35 40 45 50 6 6 12 18 24 30 36 42 48 54 60 7 7 14 21 28 35 42 49 56 63 70 8 8 16 24 32 40 48 56 64 72 80 9 9 18 27 36 45 54 63 72 81 90 10 10 20 30 40 50 60 70 80 90 100

Anatomy of the mulitplicative group Greg Martin

slide-47
SLIDE 47

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

An application: the Erd˝

  • s multiplication table problem

Question:

How many distinct integers are in the N × N multiplication table? × 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 12 14 16 18 20 3 3 6 9 12 15 18 21 24 27 30 4 4 8 12 16 20 24 28 32 36 40 5 5 10 15 20 25 30 35 40 45 50 6 6 12 18 24 30 36 42 48 54 60 7 7 14 21 28 35 42 49 56 63 70 8 8 16 24 32 40 48 56 64 72 80 9 9 18 27 36 45 54 63 72 81 90 10 10 20 30 40 50 60 70 80 90 100

Anatomy of the mulitplicative group Greg Martin

slide-48
SLIDE 48

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Spot the trend

# of distinct integers % of distinct integers N in N × N table in N × N table 10 42 42.0% 101.5 339 33.9% 102 2,906 29.0% 102.5 26,643 26.6% 103 248,083 24.8% 103.5 2,346,562 23.5% 104 22,504,348 22.5%

Time to vote:

The percentage does tend to a limit as N → ∞. Is that limit positive or zero?

Anatomy of the mulitplicative group Greg Martin

slide-49
SLIDE 49

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Spot the trend

# of distinct integers % of distinct integers N in N × N table in N × N table 10 42 42.0% 101.5 339 33.9% 102 2,906 29.0% 102.5 26,643 26.6% 103 248,083 24.8% 103.5 2,346,562 23.5% 104 22,504,348 22.5%

Time to vote:

The percentage does tend to a limit as N → ∞. Is that limit positive or zero?

Anatomy of the mulitplicative group Greg Martin

slide-50
SLIDE 50

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Table trouble

Theorem (Erd˝

  • s, 1960)

The percentage of distinct integers in the N × N multiplication table tends to 0% as N → ∞.

Four-sentence proof

Almost all integers between 1 and N have about log log N prime

  • factors. Hence almost all products in the N × N table have about

2 log log N prime factors. But almost all potential entries between 1 and N2 have only about log log(N2) ∼ log log N prime factors. Thus almost no potential entries actually appear.

Theorem (Ford, 2009)

There are about N2/(log N)δ(log log N)3/2 distinct integers in the N × N table, where δ = 1 − (1 + log log 2)/ log 2 ≈ 0.086.

Anatomy of the mulitplicative group Greg Martin

slide-51
SLIDE 51

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Table trouble

Theorem (Erd˝

  • s, 1960)

The percentage of distinct integers in the N × N multiplication table tends to 0% as N → ∞.

Four-sentence proof

Almost all integers between 1 and N have about log log N prime

  • factors. Hence almost all products in the N × N table have about

2 log log N prime factors. But almost all potential entries between 1 and N2 have only about log log(N2) ∼ log log N prime factors. Thus almost no potential entries actually appear.

Theorem (Ford, 2009)

There are about N2/(log N)δ(log log N)3/2 distinct integers in the N × N table, where δ = 1 − (1 + log log 2)/ log 2 ≈ 0.086.

Anatomy of the mulitplicative group Greg Martin

slide-52
SLIDE 52

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Table trouble

Theorem (Erd˝

  • s, 1960)

The percentage of distinct integers in the N × N multiplication table tends to 0% as N → ∞.

Four-sentence proof

Almost all integers between 1 and N have about log log N prime

  • factors. Hence almost all products in the N × N table have about

2 log log N prime factors. But almost all potential entries between 1 and N2 have only about log log(N2) ∼ log log N prime factors. Thus almost no potential entries actually appear.

Theorem (Ford, 2009)

There are about N2/(log N)δ(log log N)3/2 distinct integers in the N × N table, where δ = 1 − (1 + log log 2)/ log 2 ≈ 0.086.

Anatomy of the mulitplicative group Greg Martin

slide-53
SLIDE 53

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Table trouble

Theorem (Erd˝

  • s, 1960)

The percentage of distinct integers in the N × N multiplication table tends to 0% as N → ∞.

Four-sentence proof

Almost all integers between 1 and N have about log log N prime

  • factors. Hence almost all products in the N × N table have about

2 log log N prime factors. But almost all potential entries between 1 and N2 have only about log log(N2) ∼ log log N prime factors. Thus almost no potential entries actually appear.

Theorem (Ford, 2009)

There are about N2/(log N)δ(log log N)3/2 distinct integers in the N × N table, where δ = 1 − (1 + log log 2)/ log 2 ≈ 0.086.

Anatomy of the mulitplicative group Greg Martin

slide-54
SLIDE 54

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Table trouble

Theorem (Erd˝

  • s, 1960)

The percentage of distinct integers in the N × N multiplication table tends to 0% as N → ∞.

Four-sentence proof

Almost all integers between 1 and N have about log log N prime

  • factors. Hence almost all products in the N × N table have about

2 log log N prime factors. But almost all potential entries between 1 and N2 have only about log log(N2) ∼ log log N prime factors. Thus almost no potential entries actually appear.

Theorem (Ford, 2009)

There are about N2/(log N)δ(log log N)3/2 distinct integers in the N × N table, where δ = 1 − (1 + log log 2)/ log 2 ≈ 0.086.

Anatomy of the mulitplicative group Greg Martin

slide-55
SLIDE 55

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Table trouble

Theorem (Erd˝

  • s, 1960)

The percentage of distinct integers in the N × N multiplication table tends to 0% as N → ∞.

Four-sentence proof

Almost all integers between 1 and N have about log log N prime

  • factors. Hence almost all products in the N × N table have about

2 log log N prime factors. But almost all potential entries between 1 and N2 have only about log log(N2) ∼ log log N prime factors. Thus almost no potential entries actually appear.

Theorem (Ford, 2009)

There are about N2/(log N)δ(log log N)3/2 distinct integers in the N × N table, where δ = 1 − (1 + log log 2)/ log 2 ≈ 0.086.

Anatomy of the mulitplicative group Greg Martin

slide-56
SLIDE 56

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The largest invariant factor

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The largest invariant factor dm is the exponent of the group G (the largest order of any element). The exponent of the multiplicative group Z×

n is called the Carmichael lambda

function λ(n) (and is a divisor of φ(n)).

Theorem (Erd˝

  • s–Pomerance–Schmutz, 1991)

For almost all integers n, λ(n) ≈ n exp(log log n log log log n) = n (log n)log log log n .

Anatomy of the mulitplicative group Greg Martin

slide-57
SLIDE 57

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The largest invariant factor

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The largest invariant factor dm is the exponent of the group G (the largest order of any element). The exponent of the multiplicative group Z×

n is called the Carmichael lambda

function λ(n) (and is a divisor of φ(n)).

Theorem (Erd˝

  • s–Pomerance–Schmutz, 1991)

For almost all integers n, λ(n) ≈ n exp(log log n log log log n) = n (log n)log log log n .

Anatomy of the mulitplicative group Greg Martin

slide-58
SLIDE 58

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The largest invariant factor

Invariant factors

Every finite abelian group G is uniquely isomorphic to a direct sum of cyclic groups G ∼ = Z+

d1 ⊕ Z+ d2 ⊕ · · · ⊕ Z+ dm where

d1 | d2 | · · · | dm. The largest invariant factor dm is the exponent of the group G (the largest order of any element). The exponent of the multiplicative group Z×

n is called the Carmichael lambda

function λ(n) (and is a divisor of φ(n)).

Theorem (Erd˝

  • s–Pomerance–Schmutz, 1991)

For almost all integers n, λ(n) ≈ n exp(log log n log log log n) = n (log n)log log log n .

Anatomy of the mulitplicative group Greg Martin

slide-59
SLIDE 59

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Lambdas all the way down

One pseudorandom number generator repeatedly raises the previous number to the bth power modulo n; the period of the resulting “power-generator sequence” is a divisor of λ(λ(n)).

Theorem (M.–Pomerance, 2005)

For almost all integers n, λ(λ(n)) ≈ n exp

  • (log log n)2 log log log n

. Assuming GRH, then almost all n have at least one power-generator sequence of this length.

Higher iterates

In his PhD dissertation, Nick Harland has generalized this theorem to any higher iterate λ(λ(· · · λ(n) · · · )).

Anatomy of the mulitplicative group Greg Martin

slide-60
SLIDE 60

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Lambdas all the way down

One pseudorandom number generator repeatedly raises the previous number to the bth power modulo n; the period of the resulting “power-generator sequence” is a divisor of λ(λ(n)).

Theorem (M.–Pomerance, 2005)

For almost all integers n, λ(λ(n)) ≈ n exp

  • (log log n)2 log log log n

. Assuming GRH, then almost all n have at least one power-generator sequence of this length.

Higher iterates

In his PhD dissertation, Nick Harland has generalized this theorem to any higher iterate λ(λ(· · · λ(n) · · · )).

Anatomy of the mulitplicative group Greg Martin

slide-61
SLIDE 61

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Lambdas all the way down

One pseudorandom number generator repeatedly raises the previous number to the bth power modulo n; the period of the resulting “power-generator sequence” is a divisor of λ(λ(n)).

Theorem (M.–Pomerance, 2005)

For almost all integers n, λ(λ(n)) ≈ n exp

  • (log log n)2 log log log n

. Assuming GRH, then almost all n have at least one power-generator sequence of this length.

Higher iterates

In his PhD dissertation, Nick Harland has generalized this theorem to any higher iterate λ(λ(· · · λ(n) · · · )).

Anatomy of the mulitplicative group Greg Martin

slide-62
SLIDE 62

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Length of the primary decomposition

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ (each qj is prime).

The length ℓ is related to ω(λ(n)). (To get ℓ exactly, some primes have to be counted with multiplicity.)

Theorem (Erd˝

  • s–Pomerance, 1985)

lim

x→∞

1 x#

  • n ≤ x: ω(λ(n)) − 1

2(log log n)2

  • 1

3(log log n)3

< α

  • =

1 √ 2π α

−∞

e−t2/2 dt = Φ(α).

Anatomy of the mulitplicative group Greg Martin

slide-63
SLIDE 63

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Length of the primary decomposition

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ (each qj is prime).

The length ℓ is related to ω(λ(n)). (To get ℓ exactly, some primes have to be counted with multiplicity.)

Theorem (Erd˝

  • s–Pomerance, 1985)

lim

x→∞

1 x#

  • n ≤ x: ω(λ(n)) − 1

2(log log n)2

  • 1

3(log log n)3

< α

  • =

1 √ 2π α

−∞

e−t2/2 dt = Φ(α).

Anatomy of the mulitplicative group Greg Martin

slide-64
SLIDE 64

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Length of the primary decomposition

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ (each qj is prime).

The length ℓ is related to ω(λ(n)). (To get ℓ exactly, some primes have to be counted with multiplicity.)

Theorem (Erd˝

  • s–Pomerance, 1985)

lim

x→∞

1 x#

  • n ≤ x: ℓ − 1

2(log log n)2

  • 1

3(log log n)3

< α

  • =

1 √ 2π α

−∞

e−t2/2 dt = Φ(α).

Anatomy of the mulitplicative group Greg Martin

slide-65
SLIDE 65

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Largest primary factor

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ (qr1

1 ≤ · · · ≤ qrℓ ℓ ).

If G = Z+

n , then (almost all the time) the size of the largest

primary factor is simply P(n), the largest prime factor of n.

Theorem (Dickman–de Bruijn rho function)

The probability that P(n) is less than nα equals ρ(1/α), where

  • ρ(u) = 1,

for 0 < u ≤ 1, ρ′(u) = −ρ(u − 1)/u, for u > 1. When G = Z×

n we have heuristics and conjectures (involving

self-convolutions of ρ(u)), but the problem is still open.

Anatomy of the mulitplicative group Greg Martin

slide-66
SLIDE 66

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Largest primary factor

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ (qr1

1 ≤ · · · ≤ qrℓ ℓ ).

If G = Z+

n , then (almost all the time) the size of the largest

primary factor is simply P(n), the largest prime factor of n.

Theorem (Dickman–de Bruijn rho function)

The probability that P(n) is less than nα equals ρ(1/α), where

  • ρ(u) = 1,

for 0 < u ≤ 1, ρ′(u) = −ρ(u − 1)/u, for u > 1. When G = Z×

n we have heuristics and conjectures (involving

self-convolutions of ρ(u)), but the problem is still open.

Anatomy of the mulitplicative group Greg Martin

slide-67
SLIDE 67

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Largest primary factor

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ (qr1

1 ≤ · · · ≤ qrℓ ℓ ).

If G = Z+

n , then (almost all the time) the size of the largest

primary factor is simply P(n), the largest prime factor of n.

Theorem (Dickman–de Bruijn rho function)

The probability that P(n) is less than nα equals ρ(1/α), where

  • ρ(u) = 1,

for 0 < u ≤ 1, ρ′(u) = −ρ(u − 1)/u, for u > 1. When G = Z×

n we have heuristics and conjectures (involving

self-convolutions of ρ(u)), but the problem is still open.

Anatomy of the mulitplicative group Greg Martin

slide-68
SLIDE 68

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Largest primary factor

Primary decomposition

Every finite abelian group G is isomorphic to a direct sum of cyclic groups G ∼ = Z+

qr1

1 ⊕ Z+

qr2

2 ⊕ · · · ⊕ Z+

q

rℓ ℓ (qr1

1 ≤ · · · ≤ qrℓ ℓ ).

If G = Z+

n , then (almost all the time) the size of the largest

primary factor is simply P(n), the largest prime factor of n.

Theorem (Dickman–de Bruijn rho function)

The probability that P(n) is less than nα equals ρ(1/α), where

  • ρ(u) = 1,

for 0 < u ≤ 1, ρ′(u) = −ρ(u − 1)/u, for u > 1. When G = Z×

n we have heuristics and conjectures (involving

self-convolutions of ρ(u)), but the problem is still open.

Anatomy of the mulitplicative group Greg Martin

slide-69
SLIDE 69

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of order two

Question

How many solutions are there to x2 ≡ 1 (mod n)? Equivalently (almost), how many elements of order two are there in Z×

n ?

Answer, and average value

There are 2ω(n) solutions (mod n); and 1

x

  • n≤x 2ω(n) ∼

6 π2 log x.

Paradox

For almost all integers, 2ω(n) ≈ 2log log n = (log n)log 2. But the average value is ≈ (log x)1, which is significantly larger. So for example, when x is large, 0.1% of the integers up to x have more than 99.9% of the total number of divisors.

Anatomy of the mulitplicative group Greg Martin

slide-70
SLIDE 70

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of order two

Question

How many solutions are there to x2 ≡ 1 (mod n)? Equivalently (almost), how many elements of order two are there in Z×

n ?

Answer, and average value

There are 2ω(n) solutions (mod n); and 1

x

  • n≤x 2ω(n) ∼

6 π2 log x.

Paradox

For almost all integers, 2ω(n) ≈ 2log log n = (log n)log 2. But the average value is ≈ (log x)1, which is significantly larger. So for example, when x is large, 0.1% of the integers up to x have more than 99.9% of the total number of divisors.

Anatomy of the mulitplicative group Greg Martin

slide-71
SLIDE 71

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of order two

Question

How many solutions are there to x2 ≡ 1 (mod n)? Equivalently (almost), how many elements of order two are there in Z×

n ?

Answer, and average value

There are 2ω(n) solutions (mod n); and 1

x

  • n≤x 2ω(n) ∼

6 π2 log x.

Paradox

For almost all integers, 2ω(n) ≈ 2log log n = (log n)log 2. But the average value is ≈ (log x)1, which is significantly larger. So for example, when x is large, 0.1% of the integers up to x have more than 99.9% of the total number of divisors.

Anatomy of the mulitplicative group Greg Martin

slide-72
SLIDE 72

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of order two

Question

How many solutions are there to x2 ≡ 1 (mod n)? Equivalently (almost), how many elements of order two are there in Z×

n ?

Answer, and average value

There are 2ω(n) solutions (mod n); and 1

x

  • n≤x 2ω(n) ∼

6 π2 log x.

Paradox

For almost all integers, 2ω(n) ≈ 2log log n = (log n)log 2. But the average value is ≈ (log x)1, which is significantly larger. So for example, when x is large, 0.1% of the integers up to x have more than 99.9% of the total number of divisors.

Anatomy of the mulitplicative group Greg Martin

slide-73
SLIDE 73

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of order two

Question

How many solutions are there to x2 ≡ 1 (mod n)? Equivalently (almost), how many elements of order two are there in Z×

n ?

Answer, and average value

There are 2ω(n) solutions (mod n); and 1

x

  • n≤x 2ω(n) ∼

6 π2 log x.

Paradox

For almost all integers, 2ω(n) ≈ 2log log n = (log n)log 2. But the average value is ≈ (log x)1, which is significantly larger. So for example, when x is large, 0.1% of the integers up to x have more than 99.9% of the total number of divisors.

Anatomy of the mulitplicative group Greg Martin

slide-74
SLIDE 74

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of order two

Question

How many solutions are there to x2 ≡ 1 (mod n)? Equivalently (almost), how many elements of order two are there in Z×

n ?

Answer, and average value

There are 2ω(n) solutions (mod n); and 1

x

  • n≤x 2ω(n) ∼

6 π2 log x.

Paradox

For almost all integers, 2ω(n) ≈ 2log log n = (log n)log 2. But the average value is ≈ (log x)1, which is significantly larger. So for example, when x is large, 0.1% of the integers up to x have more than 99.9% of the total number of divisors.

Anatomy of the mulitplicative group Greg Martin

slide-75
SLIDE 75

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of order two

Question

How many solutions are there to x2 ≡ 1 (mod n)? Equivalently (almost), how many elements of order two are there in Z×

n ?

Answer, and average value

There are 2ω(n) solutions (mod n); and 1

x

  • n≤x 2ω(n) ∼

6 π2 log x.

Paradox

For almost all integers, 2ω(n) ≈ 2log log n = (log n)log 2. But the average value is ≈ (log x)1, which is significantly larger. So for example, when x is large, 0.1% of the integers up to x have more than 99.9% of the total number of divisors.

Anatomy of the mulitplicative group Greg Martin

slide-76
SLIDE 76

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of order two

Question

How many solutions are there to x2 ≡ 1 (mod n)? Equivalently (almost), how many elements of order two are there in Z×

n ?

Answer, and average value

There are 2ω(n) solutions (mod n); and 1

x

  • n≤x 2ω(n) ∼

6 π2 log x.

Paradox

For almost all integers, 2ω(n) ≈ 2log log n = (log n)log 2. But the average value is ≈ (log x)1, which is significantly larger. So for example, when x is large, 0.1% of the integers up to x have more than 99.9% of the total number of divisors.

Anatomy of the mulitplicative group Greg Martin

slide-77
SLIDE 77

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of a fixed order

The average number of elements of order 2 is

6 π2 log x.

Generalization (Finch–M.–Sebah, 2010)

There exists a constant Ck such that the average number of elements of order k in Z×

n is Ck(log x)τ(k)−1, where τ(k) is the

number of divisors of k. The same holds for the average number of solutions to xk ≡ 1 (mod n).

Variant (Finch–M.–Sebah)

The average number of solutions to xk ≡ 0 (mod n) is Dk(log x)k−1, where Dk = 1 k!(k − 1)!

  • p
  • 1 + k − 1

p

  • 1 − 1

p k−1

Anatomy of the mulitplicative group Greg Martin

slide-78
SLIDE 78

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of a fixed order

The average number of elements of order 2 is

6 π2 log x.

Generalization (Finch–M.–Sebah, 2010)

There exists a constant Ck such that the average number of elements of order k in Z×

n is Ck(log x)τ(k)−1, where τ(k) is the

number of divisors of k. The same holds for the average number of solutions to xk ≡ 1 (mod n).

Variant (Finch–M.–Sebah)

The average number of solutions to xk ≡ 0 (mod n) is Dk(log x)k−1, where Dk = 1 k!(k − 1)!

  • p
  • 1 + k − 1

p

  • 1 − 1

p k−1

Anatomy of the mulitplicative group Greg Martin

slide-79
SLIDE 79

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of a fixed order

The average number of elements of order 2 is

6 π2 log x.

Generalization (Finch–M.–Sebah, 2010)

There exists a constant Ck such that the average number of elements of order k in Z×

n is Ck(log x)τ(k)−1, where τ(k) is the

number of divisors of k. The same holds for the average number of solutions to xk ≡ 1 (mod n).

Variant (Finch–M.–Sebah)

The average number of solutions to xk ≡ 0 (mod n) is Dk(log x)k−1, where Dk = 1 k!(k − 1)!

  • p
  • 1 + k − 1

p

  • 1 − 1

p k−1

Anatomy of the mulitplicative group Greg Martin

slide-80
SLIDE 80

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Elements of a fixed order

The average number of elements of order 2 is

6 π2 log x.

Generalization (Finch–M.–Sebah, 2010)

There exists a constant Ck such that the average number of elements of order k in Z×

n is Ck(log x)τ(k)−1, where τ(k) is the

number of divisors of k. The same holds for the average number of solutions to xk ≡ 1 (mod n).

Variant (Finch–M.–Sebah)

The average number of solutions to xk ≡ 0 (mod n) is Dk(log x)k−1, where Dk = 1 k!(k − 1)!

  • p
  • 1 + k − 1

p

  • 1 − 1

p k−1

Anatomy of the mulitplicative group Greg Martin

slide-81
SLIDE 81

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Gory details: The constant Ck

Roots of unity

Average number of solutions to xk ≡ 1 (mod n) is Ck(log x)τ(k)−1 Ck = θ(k) (τ(k) − 1)!

  • p
  • 1 + (k, p − 1)

p − 1

  • 1 − 1

p τ(k) where θ(k) is defined as follows: if k = 2ik0 with k0 odd, then θ(k) =

  • 1,

if i = 0, (i + 5)/4, if i ≥ 1

pjk0

  • 1 +

j(k, p − 1)(p − 1) p(p + (k, p − 1) − 1)

  • Anatomy of the mulitplicative group

Greg Martin

slide-82
SLIDE 82

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of subgroups

Definition

Let Gn denote the number of subgroups of Z×

n (as sets, not up

to isomorphism).

How big can Gn get?

Anatomy of the mulitplicative group Greg Martin

slide-83
SLIDE 83

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of subgroups

Definition

Let Gn denote the number of subgroups of Z×

n (as sets, not up

to isomorphism).

How big can Gn get?

Anatomy of the mulitplicative group Greg Martin

slide-84
SLIDE 84

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of subgroups

Definition

Let Gn denote the number of subgroups of Z×

n (as sets, not up

to isomorphism).

How big can Gn get?

> log n

Anatomy of the mulitplicative group Greg Martin

slide-85
SLIDE 85

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of subgroups

Definition

Let Gn denote the number of subgroups of Z×

n (as sets, not up

to isomorphism).

How big can Gn get?

> log n > τ(n), which at its largest is ≈ n(log 2)/ log log n

Anatomy of the mulitplicative group Greg Martin

slide-86
SLIDE 86

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of subgroups

Definition

Let Gn denote the number of subgroups of Z×

n (as sets, not up

to isomorphism).

How big can Gn get?

> log n > τ(n), which at its largest is ≈ n(log 2)/ log log n > φ(n), which is larger than ≈ n/(log log n)

Anatomy of the mulitplicative group Greg Martin

slide-87
SLIDE 87

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of subgroups

Definition

Let Gn denote the number of subgroups of Z×

n (as sets, not up

to isomorphism).

How big can Gn get?

> log n > τ(n), which at its largest is ≈ n(log 2)/ log log n > φ(n), which is larger than ≈ n/(log log n) > n10100

Anatomy of the mulitplicative group Greg Martin

slide-88
SLIDE 88

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of subgroups

Definition

Let Gn denote the number of subgroups of Z×

n (as sets, not up

to isomorphism).

How big can Gn get?

> log n > τ(n), which at its largest is ≈ n(log 2)/ log log n > φ(n), which is larger than ≈ n/(log log n) > n10100

There are infinitely many n . . .

. . . for which Gn > exp

  • c(log n)2/(log log n)2

. . . even if we count only subgroups that look like Z2 ⊕ · · · ⊕ Z2!

Anatomy of the mulitplicative group Greg Martin

slide-89
SLIDE 89

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The number of subgroups

Definition

Let Gn denote the number of subgroups of Z×

n (as sets, not up

to isomorphism).

How big can Gn get?

> log n > τ(n), which at its largest is ≈ n(log 2)/ log log n > φ(n), which is larger than ≈ n/(log log n) > n10100

There are infinitely many n . . .

. . . for which Gn > exp

  • c(log n)2/(log log n)2

. . . even if we count only subgroups that look like Z2 ⊕ · · · ⊕ Z2!

Anatomy of the mulitplicative group Greg Martin

slide-90
SLIDE 90

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Relating Gn to additive functions

G(n) = number of subgroups of Z×

n

Notation

Let ωq(n) denote the number of primes dividing n that are congruent to 1 (mod q).

A sum of squares of additive functions

One can show: log Gn ≈ 1

4

  • ω2(n)2 + ω3(n)2 + ω4(n)2 + ω5(n)2

+ ω7(n)2 + ω8(n)2 + ω9(n)2 + ω11(n)2 + · · ·

  • = 1

4

  • pr ωpr(n)2.

Anatomy of the mulitplicative group Greg Martin

slide-91
SLIDE 91

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Relating Gn to additive functions

G(n) = number of subgroups of Z×

n

Notation

Let ωq(n) denote the number of primes dividing n that are congruent to 1 (mod q).

A sum of squares of additive functions

One can show: log Gn ≈ 1

4

  • ω2(n)2 + ω3(n)2 + ω4(n)2 + ω5(n)2

+ ω7(n)2 + ω8(n)2 + ω9(n)2 + ω11(n)2 + · · ·

  • = 1

4

  • pr ωpr(n)2.

Anatomy of the mulitplicative group Greg Martin

slide-92
SLIDE 92

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Relating Gn to additive functions

G(n) = number of subgroups of Z×

n

Notation

Let ωq(n) denote the number of primes dividing n that are congruent to 1 (mod q).

A sum of squares of additive functions

One can show: log Gn ≈ 1

4

  • ω2(n)2 + ω3(n)2 + ω4(n)2 + ω5(n)2

+ ω7(n)2 + ω8(n)2 + ω9(n)2 + ω11(n)2 + · · ·

  • = 1

4

  • pr ωpr(n)2.

Anatomy of the mulitplicative group Greg Martin

slide-93
SLIDE 93

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Aspirations

G(n) = number of subgroups of Z×

n

Work currently in progress

I plan to establish an Erd˝

  • s–Kac-type theorem demonstrating a

Gaussian distribution not just for additive functions, but for products of additive functions, and sums of such products.

Hopeful theorem

I believe I can show: lim

x→∞

1 x#

  • n ≤ x: log Gn − A(log log n)2 < α
  • B(log log n)3
  • =

1 √ 2π α

−∞

e−t2/2 dt = Φ(α).

Anatomy of the mulitplicative group Greg Martin

slide-94
SLIDE 94

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Aspirations

G(n) = number of subgroups of Z×

n

Work currently in progress

I plan to establish an Erd˝

  • s–Kac-type theorem demonstrating a

Gaussian distribution not just for additive functions, but for products of additive functions, and sums of such products.

Hopeful theorem

I believe I can show: lim

x→∞

1 x#

  • n ≤ x: log Gn − A(log log n)2 < α
  • B(log log n)3
  • =

1 √ 2π α

−∞

e−t2/2 dt = Φ(α).

Anatomy of the mulitplicative group Greg Martin

slide-95
SLIDE 95

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

Gory details: The constants A and B

Hopeful theorem

lim

x→∞

1 x#

  • n ≤ x: log Gn − A(log log n)2 < α
  • B(log log n)3
  • =

1 √ 2π α

−∞

e−t2/2 dt = Φ(α) A = 1

4

  • p

p2 log p (p − 1)3(p + 1) ≈ 0.374516 B = 4A2 + 1

4

  • p

p3(p4 − p3 − p2 − p − 1)(log p)2 (p − 1)6(p + 1)2(p2 + p + 1) ≈ 0.617393

Anatomy of the mulitplicative group Greg Martin

slide-96
SLIDE 96

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The end

These slides

www.math.ubc.ca/∼gerg/index.shtml?slides

My paper with Pomerance on λ(λ(n))

www.math.ubc.ca/∼gerg/ index.shtml?abstract=ICFNCPG

My paper with Finch and Sebah on roots of 1 and 0

www.math.ubc.ca/∼gerg/ index.shtml?abstract=RUNM

Anatomy of the mulitplicative group Greg Martin

slide-97
SLIDE 97

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The end

These slides

www.math.ubc.ca/∼gerg/index.shtml?slides

My paper with Pomerance on λ(λ(n))

www.math.ubc.ca/∼gerg/ index.shtml?abstract=ICFNCPG

My paper with Finch and Sebah on roots of 1 and 0

www.math.ubc.ca/∼gerg/ index.shtml?abstract=RUNM

My papers on products of additive functions and subgroups of Z×

n

Keep an eye on the arXiv!

Anatomy of the mulitplicative group Greg Martin

slide-98
SLIDE 98

Questions about Z×

n

Distribution of ω(n) Other invariants of Z×

n

Elements and subgroups

The end

These slides

www.math.ubc.ca/∼gerg/index.shtml?slides

My paper with Pomerance on λ(λ(n))

www.math.ubc.ca/∼gerg/ index.shtml?abstract=ICFNCPG

My paper with Finch and Sebah on roots of 1 and 0

www.math.ubc.ca/∼gerg/ index.shtml?abstract=RUNM

My papers on products of additive functions and subgroups of Z×

n

Keep an eye on the arXiv! (but don’t hold your breath. . . )

Anatomy of the mulitplicative group Greg Martin