MaBloWriMo 16: Recap and outline

We have now established all the facts we will need about groups, and have incidentally just passed the halfway point of MaBloWriMo. This feels like a good time to take a step back and outline what we’ve done so far and where we are going.

So far:

  • We defined s_0 = 4 and s_n = s_{n-1}^2 - 2; the Lucas-Lehmer test says that M_n = 2^n - 1 is prime if and only if s_{n-2} is divisible by M_n. Currently, we’re trying to prove the backwards direction: if s_{n-2} is divisible by M_n, then M_n is prime.
  • We defined \omega = 2 + \sqrt 3 and \overline{\omega} = 2 - \sqrt  3, and proved that \omega \overline{\omega} = 1, and s_m =  \omega^{2^m} + \overline{\omega}^{2^m}.
  • We learned the definition of groups, looked at some examples, and proved some simple facts, such as:
    • Every element of a finite group has a finite order.
    • The order of an element is at most the size of the group.
    • If g^n = e then the order of g divides n.

We’re now going to start the proof proper, which will be a proof by contradiction. So we will assume that s_{n-2} is divisible by M_n, but M_n is not prime. From there:

  • We will define a group that contains \omega and \overline{\omega} as elements. The group will be defined in terms of a nontrivial divisor of M_n.
  • Using the facts we proved about groups, and the fact that M_n divides s_{n-2}, we will show that the order of \omega has to be 2^n.
  • Finally, we will show that the order of the group has to be less than 2^n—a contradiction, since the order of elements is never greater than the order of the group.

Tomorrow: we’ll start in on defining the crucial group that contains \omega.

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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2 Responses to MaBloWriMo 16: Recap and outline

  1. Pingback: MaBloWriMo 20: the group X star | The Math Less Traveled

  2. Pingback: MaBloWriMo 21: the order of omega, part I | The Math Less Traveled

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