2 consider the unit group g u2800 a find the order of the...

2. Consider the unit group $G = U_{2800}$. a. Find the order of the group $U_{2800}$. b. Prove that the only solution to $x^{77} - \bar{1} = 0$ in $\mathbb{Z}/2800\mathbb{Z}$ is $x = \bar{1}$. Suggestion. Use the results of problem 1.

Transcript

00:01 In this problem, we're given the group g equals z27 star with respect to multiplication, the group of units, modulo 27.

00:09 Our first job is to find the order of the group.

00:12 Our second job is to find the number of primitive roots of the group.

00:15 Then our third job is to list those primitive roots.

00:18 So we will start by finding the order of this group.

00:26 We know that the order of such a group is given.

00:30 We use little o for order.

00:32 Order of the order of g is given by taking the euler phi function and applying it to 27.

00:39 So we will factor 27 into its prime decomposition.

00:42 It is 3 to the third.

00:44 So this function value is 3 to the third minus 3 squared.

00:50 So we get 27 minus 9, and we see that this group contains 18 elements in all.

00:57 So there's our answer for the first part.

01:02 For the second part, we we know that 27 has the proper form for there to be primitive roots at all.

01:12 It is a prime power.

01:13 It has to be an odd prime power or two times an odd prime power or two or four.

01:18 So the number of primitive roots, we'll just abbreviate those as pr.

01:24 The number of primitive roots is phi of phi of 27.

01:31 So this means we need now to calculate phi of 18 using our answer from part a.

01:38 So we'll factor 18 into relatively prime factors, 2 times 3 squared, use the fact that the order of the function is multiplicative, so we get 1 times 3 squared, that's a 2, sorry about that, 3 squared minus 3 to the first, and so we get 9 minus 3 or 6 primitive roots in all.

02:06 So there's our answer for the second part.

02:11 And so to find these six primitive roots in part c, we will start with one of the elements in the group.

02:22 We will start with two.

02:23 Of course, one has order one, so we will generate the subgroup using two.

02:28 We will generate its cyclic subgroup.

02:30 We're just going to look at powers of two and reduce those modulo 27 as we go.

02:35 So these powers start out two, four, eight, 16.

02:41 And of course when we get to the next power, 32 reduces to 5 mod 27...