Math 417 – Seventh Day – Class, Lecture notes of Algebra

Download Math 417 – Seventh Day – Class and more Lecture notes Algebra in PDF only on Docsity!

Math 417 – Seventh Day – Class

Bruce Reznick University of Illinois at Urbana-Champaign

September 9, 2020

First a few elaborations on the material I sent out Tuesday.

Given that n ≡ 1 , 3 , 7 , 9 mod 10, it follows that [n] 10 ∈ (Z/ 10 Z)∗, so gcd(n, 10) = 1 = gcd(10, n), so that [10]n ∈ (Z/nZ)∗.

Rather than appealing to a previous theorem, I’ll give the proof again. We know that (Z/nZ)∗^ is a finite set, so the sequence

[10]n, [10^2 ]n, [10^3 ]n, [10^4 ]n,...

can’t all be different.

Thus, we can find i < j so that [10i^ ]n = [10j^ ]n. (This is kind of the Pigeonhole Principle.) Thus, since gcd(n, 10) = 1,

n | 10 j^ − 10 i^ = 10i^ (10j−i^ − 1) =⇒ n | (10j−i^ − 1) =⇒ 10 j−i^ ≡ 1 mod n =⇒ [10j−i^ ]n = [1]n.

For example, if n = 13, then Mathematica tells me that the sequence of 10k^ mod 13 is:

Here, i = 0, j = 6, and 10^6 − 1 = 999999 = 13 · 76923.

One more question was about the order of composition, and I think it would be easiest if I used two functions which aren’t permutations.

Suppose f (x) = x + 3 and g (x) = x^2 and you apply f first and then do g. Then you get x 7 → x + 3 7 → (x + 3)^2. For comparison,

f (g (x)) = f (x^2 ) = x^2 + 3, g (f (x)) = g (x + 3) = (x + 3)^2.

So in this case too: if we act in the order written, it gets written in the reverse. It’s confusing but this is how it works here.

The symmetric group we will get to know best is S 3 , which has 6 = 3! elements. I will write them down and then give them all names (the same ones in Fraleigh) and say a little bit about each one. They can also be viewed as symmetries of an equilateral triangle.

Here is the triangle:

1 2

Each element of S 3 gets its own page.

This is the second element, ρ 1. Under ρ 1 ,

1 7 → 2 , 2 7 → 3 , 3 7 → 1

ρ 1 =

The permutation ρ 1 rotates the triangle clockwise by 23 π. The triangle shows that the label 1 goes to the position 2, the label 2 goes to the position 3 and label 3 goes to the position 1.

This is the third element, ρ 2. Under ρ 2 ,

1 7 → 3 , 2 7 → 1 , 3 7 → 2

ρ 2 =

The permutation ρ 2 rotates clockwise by 43 π or counterclockwise by 2 π

3. The triangle shows that the label 1 goes to the position 3, the label 2 goes to the position 1 and label 3 goes to the position 2.

I hope you can see that ρ 2 = ρ^21 and ρ 1 = ρ^22 , ρ^31 = ρ^32 = ρ 0 = e. It follows that {ρ 0 , ρ 1 , ρ 2 } is a cyclic group of order 3, and a subgroup of S 3.

This is the fifth element, μ 2. Under μ 2 ,

1 7 → 3 , 2 7 → 2 , 3 7 → 1

μ 2 =

The permutation μ 2 flips 1 and 3 and fixes 2. Also a transposition. Notice that μ^22 = ρ 0 = e. It’s equivalent to flipping the triangle on a diameter through 2.

This is the sixth element, μ 3. Under μ 3 ,

1 7 → 2 , 2 7 → 1 , 3 7 → 3

μ 3 =

The permutation μ 3 flips 1 and 2 and fixes 3. Also a transposition. Again, μ^23 = ρ 0 = e. It’s equivalent to flipping the triangle on a diameter through 3.

One more point on these before we go on. Imagine that the triangle is, say, red on the front and blue on the back. We start with it as red. Any rotation (or the identity) (that is, any ρi ) will keep it red. Any flip (that is, any μi ) will switch it so the front is blue.

If you think about that, then you can convince yourself, without any calculation, that ρi ◦ ρj will be some ρk and ρi ◦ μj or μi ◦ ρj will be some μk. Finally, two flips will turn the triangle twice, so it’s in its original color position. Thus, μi ◦ μj will be some ρk. (We saw that twice above.) This shows up in table 8.8 on p.79 of the book. We’ll talk about this on Friday.

WORKSHEET PROBLEMS

1. It is a fact from arithmetic that 77 · 12987 = 999999. Use this information to give the decimal expansion of

50 77

2. Recall that

ρ 2 =

= (132), μ 2 =

Calculate ρ 2 ◦ μ 2 and μ 2 ◦ ρ 2.

2. Again, for reference,

ρ 2 =

= (132), μ 2 =

So ρ 2 ◦ μ 2 takes 1 7 → 3 7 → 1, 2 7 → 1 7 → 3 and 3 7 → 2 7 → 2 and

ρ 2 ◦ μ 2 =

Similarly, μ 2 ◦ ρ 2 takes 1 7 → 3 7 → 2, 2 7 → 2 7 → 1 and 3 7 → 1 7 → 3 and

μ 2 ◦ ρ 2 =

That is, ρ 2 ◦ μ 2 = μ 1 and μ 2 ◦ ρ 2 = μ 3.