Why the need for Lagrange's Theorem?

  • Context: Graduate 
  • Thread starter Thread starter Spartan Math
  • Start date Start date
  • Tags Tags
    Theorem
Click For Summary

Discussion Overview

The discussion revolves around the necessity of Lagrange's theorem in proving that a map defined by raising elements of a finite group to a power is surjective, particularly when the power is an integer relatively prime to the group's order. The scope includes theoretical aspects of group theory and the application of Lagrange's theorem in this context.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the relevance of Lagrange's theorem in their proof regarding the surjectivity of the map x --> x^k.
  • Another participant seeks clarification on the condition x^n = 1 and its relation to the order of the group.
  • A participant acknowledges the need for Lagrange's theorem to establish that x^n = 1, but is uncertain if this is the only application of the theorem in the problem.
  • There is a suggestion that the order of the subgroup generated by x^k must divide the order of the group, although the reasoning behind this is not fully understood by all participants.

Areas of Agreement / Disagreement

Participants express uncertainty about the necessity and application of Lagrange's theorem, indicating that multiple competing views remain regarding its role in the proof.

Contextual Notes

There are unresolved assumptions about the implications of the order of elements and the subgroup generated by x^k, as well as the specific conditions under which Lagrange's theorem applies.

Spartan Math
Messages
23
Reaction score
0
I've been working on a problem and I can't seem to figure out as to why the book wants you to use Lagrange's theorem. The question is "Prove that if k is an integer that is relatively prime to the order of a finite group, then the map x --> x^k is surjective." My idea was that since n and k are relatively prime, then ns + kt = 1 so that x = x^(ns+kt) which simplifies to x = x^(ks) so that x is the image of x^s and it is surjective. So where does Lagrange's theorem come into play?
 
Last edited:
Physics news on Phys.org
How do you know that x^n=1?
 
oh, hah, sorry. I meant to say that the order of G is n, so that makes sense now.
 
Just checking: does the need for Lagrange make sense now?
 
well, not really. I'm unsure about it. Something about the order of the subgroup generated by x^k dividing the order of group, though I'm not sure why we need that.
 
OK, let me rephrase me question then: how do you know that x^n=1 without using Lagrange's theorem?
 
well, I wasn't sure if it was something that was already proved (x^(order of G) = 1). Which is what I was thinking, but yes, I know that Lagrange's theorem is necessary for "x^n = 1", though would this be the only way that we would use Lagrange's theorem in this problem?

ah, so it makes perfect sense now why they say use lagrange's theorem. I was definitely taking this for granted since I had been doing lots of algebra before this and it was probably something I found they had covered, since I haven't felt the need to read through the book.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K