Why the need for Lagrange's Theorem?

1. Sep 21, 2008

Spartan Math

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: Sep 21, 2008
2. Sep 21, 2008

morphism

How do you know that x^n=1?

3. Sep 21, 2008

Spartan Math

oh, hah, sorry. I meant to say that the order of G is n, so that makes sense now.

4. Sep 21, 2008

morphism

Just checking: does the need for Lagrange make sense now?

5. Sep 21, 2008

Spartan Math

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.

6. Sep 21, 2008

morphism

OK, let me rephrase me question then: how do you know that x^n=1 without using Lagrange's theorem?

7. Sep 21, 2008

Spartan Math

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.