Math 417 – Seventh Day, Summaries of Number Theory

Download Math 417 – Seventh Day and more Summaries Number Theory in PDF only on Docsity!

Math 417 – Seventh Day

Bruce Reznick University of Illinois at Urbana-Champaign

September 8, 2020

Welcome back. I hope you had a restful break.

Because of the technical problems in Friday’s Zoom, I want to start with the worksheet problems.

I will continue with a few topics that I neglected to mention from sections four through six and give one more application from number theory.

I will skip section seven and move on to section eight, with permutations which give some non-abelian groups.

2. Same situation. Consider C 6 = 〈a〉, a^6 = e.

There is an isomorphism Φ which takes C 6 to Z/ 2 Z × Z/ 3 Z and for which Φ(a) = ([1] 2 , [1] 3 ).

Write out the other values of Φ(ak^ ), and Φ(〈a^2 〉) and Φ(〈a^3 〉).

These are subgroups of Z/ 2 Z × Z/ 3 Z that will be “obviously” subgroups.

SOLUTION to 1. From the definition of the group:

([1] 2 , [1] 3 )^1 = ([1] 2 , [1] 3 ) ([1] 2 , [1] 3 )^2 = ([1] 2 , [1] 3 ) ∗ ([1] 2 , [1] 3 ) = ([0] 2 , [2] 3 ) ([1] 2 , [1] 3 )^3 = ([1] 2 , [1] 3 ) ∗ ([1] 2 , [1] 3 )^2 = ([1] 2 , [0] 3 ) ([1] 2 , [1] 3 )^4 = ([1] 2 , [1] 3 ) ∗ ([1] 2 , [1] 3 )^3 = ([0] 2 , [1] 3 ) ([1] 2 , [1] 3 )^5 = ([1] 2 , [1] 3 ) ∗ ([1] 2 , [1] 3 )^4 = ([1] 2 , [2] 3 ) ([1] 2 , [1] 3 )^6 = ([1] 2 , [1] 3 ) ∗ ([1] 2 , [1] 3 )^5 = ([0] 2 , [0] 3 )

So the first five powers are different and the sixth gives the identity and Z/ 2 Z × Z/ 3 Z is a cyclic group of order 6.

A couple of points worth mentioning from the book. The author uses a′^ where I have used a−^1. I don’t know why. He also has a slightly different definition of 〈a〉 than I do. The definitions coincide for finite groups (which is what we’ll mostly study), but are subtly different for infinite groups.

Recall that I defined

〈a〉 = {a, a^2 , a^3 ,... } = {an^ | n ∈ N}

Fraleigh says that

〈a〉 = {... a−^2 , a−^1 , e, a, a^2 , a^3 ,... } = {an^ | n ∈ Z}.

Suppose G is finite, then there exists m so that am^ = e, hence e ∈ 〈a〉 by my definition and, since am−^1 = a−^1 , a−^1 ∈ 〈a〉 as well, the two definitions coincide.

However, when G is an infinite group, then they can be different.

Consider (Z, +). In this case, my definition gives

〈 1 〉 = { 1 , 2 , 3 ,... } = N

Clearly, we never get to the identity – 0 – this way, and this subset is not a subgroup. Fraleigh uses the more standard and correct definition so that

〈 1 〉 = {− 2 , − 1 , 0 , 1 , 2 , 3 ,... } = Z.

He also phrases this as saying that 〈a〉 is the “smallest subgroup” of G containing a. The reason for this is twofold.

First of all, 〈a〉 = {an^ | n ∈ Z} is a group and it is a subgroup of G. Second, if H is a subgroup of G and a ∈ H, so a ∗ a = a^2 ∈ H, etc. But also e ∈ H and a−^1 ∈ H because it’s a subgroup, and so by an easy induction an^ ∈ H for n ∈ Z. Thus 〈a〉 ⊆ H, so 〈a〉 is the smallest subgroup.

= 142857 × (.000001 000001 000001 000001... )

= 142857 ×

= 142857 × 1

×

142857 × 1

× 1

=^142857

=^142857

Finally, 999999 = 7 · 142857, so the sum is 142857 999999 =^

7 ∗ 142857 =^

I was using the geometric series, but | 1016 | < 1, so that’s ok.

Here’s a wonderful fact that we are about to prove:

Suppose n ≡ 1 , 3 , 7 , 9 mod 10 and cn ∈ (0, 1), then the decimal expression for cn always repeats! For example,

The block that repeats has 46 digits. The proof of the wonderful fact is not that hard and will follow from the group theory we’ve been doing.

LEMMA 2: For any r ∈ N,

1 10 r^ +^

102 r^ +^

103 r^ +^

104 r^ +^ · · ·^ =^

10 r^ − 1.

PROOF: As we saw above with r = 6,

1 10 r^ +^

102 r^ +^

103 r^ +^

104 r^ +^ · · ·^ = 1 10 r^

10 r^

102 r^

103 r^

10 r^ ·

∑^ ∞

k=

(10r^ )k^ =^

10 r^ ·^

1 − (^101) r

10 r^ − 1.

Again, I was using the geometric series, but | (^101) r | < 1, so it’s valid.

THEOREM: Suppose n ≡ 1 , 3 , 7 , 9 mod 10 and cn ∈ (0, 1), then there exists r so that the decimal expansion of cn will repeat in a block of size r.

PROOF: By Lemma 1, there exists r so that 10r^ ≡ 1 mod n. This means that n | 10 r^ − 1 ⇐⇒ 10 r^ − 1 = n · t for some integer t. Therefore, by Lemma 2,

c n

= ct nt

= ct 10 r^ − 1

= ct · (. 00... 01 00... 01 00... 01... )

Since cn < 1, we have ct < nt = 10r^ − 1 and this means that there is no carryover when we multiply into the block.

We have 10^46 ≡ 1 mod 417, which explains the blocks of 46 above, and also φ(417) = φ(3 · 139) = (3 − 1)(139 − 1) = 2 · 138 = 276 = 6 · 46 and

1046 − 1 = 417×

Also, you can get these fractions in many different ways. You probably know that 111 has a nice decimal expression

So we can think of it in two ways: 1 11 =^

99 =^

102 − 1 = 9^ ·^ (.01 01 01 01^...^ )

Now a complete change of topics. Let A = {a 1 ,... , an} be a finite set, and let σ be a bijection of A to A. That is

{σ(a 1 ), σ(a 2 ),... , σ(an)} = {a 1 ,... , an} = A.

We say that σ is a permutation of A.

At the beginning at least, we will take A = { 1 ,... n}. As an example, if n = 5, then we might have

σ(1) = 3, σ(2) = 2, σ(3) = 5, σ(4) = 1, σ(5) = 4.

There are two ways to write this in a compressed form. The first is to consider a kind of matrix in which j lies above σ(j). In this case

σ =

As you probably suspect, there is a group at work here. We need to figure out how to combine them, and the order can be confusing.

Let me give another permutation of { 1 , 2 , 3 , 4 , 5 }

ρ(1) = 4, ρ(2) = 3, ρ(3) = 5, ρ(4) = 1, ρ(5) = 2.

so we have ρ =

Since ρ acts as follows

1 7 → 4 7 → 1 , 2 7 → 3 7 → 5 7 → 2 ,

we’d write ρ = (14)(235) = (352)(41), etc.

Since

σ =

, ρ =

we define σ ◦ ρ by reading from left to right as it acts on the various elements

σ(1) = 3 ρ(3) = 5 =⇒ (σ ◦ ρ)(1) = 5, 1 7 → 3 7 → 5; σ(2) = 2 ρ(2) = 3 =⇒ (σ ◦ ρ)(2) = 3, 2 7 → 2 7 → 3; σ(3) = 5 ρ(5) = 2 =⇒ (σ ◦ ρ)(3) = 2, 3 7 → 5 7 → 2; σ(4) = 1 ρ(1) = 4 =⇒ (σ ◦ ρ)(4) = 4, 4 7 → 1 7 → 4; σ(5) = 4 ρ(4) = 1 =⇒ (σ ◦ ρ)(5) = 1, 5 7 → 4 7 → 1.

Thus σ ◦ ρ =