Compute the orders of the following groups. a. U(3), U(4), U(12) b. U(5), U(7), U(35) c. U(4), U

Compute the orders of the following groups. a. U(3), U(4),...

Key Concepts

Multiplicative Property of the Totient Function

A key property of Euler's totient function is its multiplicative nature: if two positive integers r and s are coprime, then ?(rs) = ?(r)·?(s). This property implies that the order of U(rs) equals the product of the orders of U(r) and U(s), establishing a relationship among these groups.

Chinese Remainder Theorem

The Chinese Remainder Theorem states that if two integers r and s are coprime, then the ring of integers modulo rs is isomorphic to the direct product of the rings modulo r and s. This theorem not only provides a structural insight into modular arithmetic but also underpins the isomorphism U(rs) ? U(r) × U(s), which is closely related to the multiplicative property of the Euler totient function.

Group of Units

The group of units modulo n, denoted U(n), is the set of all integers between 1 and n that are relatively prime to n, equipped with the operation of multiplication modulo n. This group is fundamental in number theory and abstract algebra as it provides insight into the structure and properties of modular arithmetic.

Euler Totient Function

Euler's totient function, ?(n), counts the number of positive integers up to n that are coprime with n. It directly gives the order of the group of units U(n) and plays a crucial role in many results in number theory, including Euler's theorem.

Transcript

00:01 To solve this problem, we will be performing the series of composition of the relation r with itself.

00:09 But first, we will represent the regulation using the 0 -1 matrix.

00:17 So what we will get here is a 5x5 matrix.

00:26 So for the first rule, there's only 1 -3, so it will be 0 -0 -1 -0.

00:35 Zero second line second rule there's only two four so zero zero one zero here in the third line there's three one and three five so we have one here and fourth rule only the third position has a one and for the fifth rule the first the second and the fourth position have one so this is the zero on matrix here nor in part a we'll find in our billion.

01:32 So for this, this would be the composition of r with itself, which means we're going to do the boolean product of our matrix m sub r with itself.

01:54 When we multiply those out, we would do the boolean multiplication.

02:00 We end up with a matrix.

02:01 I put this here in red.

02:16 We get the matrix 1 -0 -0 -1.

02:26 Here is now 1 -0.

02:30 I'm assuming here you already know how to do the boolean product.

02:36 You can review section 2 .6 in your book if you do not remember.

02:46 1 is 0 .0.

02:58 So based on this matrix, then our square on our first row we got the points one one and one five and the second rule we have the point two three on the third rule we get three one three two three three and three four on the fourth row we have four one and four five and and the final rule, we have 5 .4 and 5.

04:26 Now in part b, we are finding ours cubed, and that would be the composition of r with our, or the boolean product of our answer in part 1, part a, sorry.

05:04 So let's call it squared here.

05:07 Let's square here.

05:08 Most of our.

05:20 So once we actually do the bullion product, what we get for our cute on the first row of the matrix, we have 1 -1, 1 -2, 1 -3, and 1 -4.

05:53 On the second row, we have 2 -1 and 2 -5.

06:05 On the third row, we find 3 -1, 3 -3, 3 -3, 3...

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