Why is the number of fixed points divisible by p in Cauchy's Theorem proof?

In summary, the author is going through a proof of Cauchy's theorem in a textbook and is stuck on a particular point. They have tried looking at other proofs, but they all use the Orbit-Stabilizer Theorem or induction. The proof presented in the book probably uses the O-S Theorem, but not directly (since it's not been covered yet). Finally, the author is able to deduce that if g^p=1, then (g,...,g) is in A and is fixed by \pi, and that all other elements of A belong to cycles of size p.
  • #1
Bleys
74
0
I'm going through the proof of Cauchy's Theorem in a textbook and I'm stuck on a particular point. I've tried looking at other proofs, but they all use the Orbit-Stabilizer Theorem or induction. The proof presented in the book probably uses the O-S Theorem, but not directly (since it's not been covered yet).

Let G be a finite group and p a prime dividing the order of G. Define the set [itex] A= \{ (g_{1},...,g_{p} ) : g_{i} \in G , g_{1}...g_{p} = 1 \} [/itex] and a permutation [itex] \pi [/itex] on A by [itex] \pi (g_{1},...,g_{p}) = (g_{2},...,g_{p},g_{1}) [/itex]
It shows that if [itex] g^p=1[/itex], then (g,...,g) is in A and is fixed by [itex]\pi[/itex], and that all other elements of A belong to cycles of size p. That's all fine. Then it shows [itex]|A| = |G|^{p-1} [/itex] and so p divides |A|. Since all cycles have size 1 or p, the number of fixed points is also divisible by p. I don't understand this. Why does the fact that |A| = pk imply that the number of fixed points is also a multiple of p?

All I've been able to deduce is that [itex] kp = |A| = an_{1} + bn_{2} [/itex] where [itex] n_{1}, n_{2}[/itex] is the number of 1 cycles and p cycles, respectively.
Could someone clear this up for me, please?

(also in LaTeX, why are \pi and n_{1},n_{2} not aligned with the rest of the line, and the curly brackets not showing up in the code for the definition of A? EDIT: fixed now, thank you Gregg!)
 
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  • #2
Bleys said:
(also in LaTeX, why are \pi and n_{1},n_{2} not aligned with the rest of the line, and the curly brackets not showing up in the code for the definition of A?)


use \{

[tex] A = \{ x : x^2=1\} [/tex]

second, use itex tags instead of tex:

I like to talk about the number[itex]\pi [/itex] during a sentence.
 
  • #3
Bleys said:
...Then it shows [itex]|A| = |G|^{p-1} [/itex] and so p divides |A|. Since all cycles have size 1 or p, the number of fixed points is also divisible by p. I don't understand this. Why does the fact that |A| = pk imply that the number of fixed points is also a multiple of p?
...
The sequences g1,g2,...,gp (if any) where g1g2...gp=e and it isn't the case that g1=g2=...=gp fall into sets of p that are cyclic permutations of each other, because if g1g2...gp=e then g2g3...gpg1=e etc.

If we remove these from A, since |A| = pk that leaves some multiple m=rp of p sequences ggg...g of elements such that gp=e. One of these is eee...e, so r is not zero. Therefore there must be rp-1>0 elements of order p.

(Incidentally these fall into subgroups of order p that intersect only in the identity and each contains p-1 elements of order p, so you can also say rp-1=0 (p-1) or r=1 (p-1), so that the number of elements of order p is actually
(h(p-1)+1)p-1=(hp+1)(p-1) for some integer h.)

By the way, I believe this is McKay's proof of Cauchy's theorem , first published in American Mathematical Monthly, Vol.66 (1959), page 119.
 
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  • #4
...fall into sets of p that are cyclic permutations of each other...
I'm assuming you mean sets of order p. Ok, it was not obvious to me that the set of non-fixed was also divisible by p (but I see how the result follows from this).
Thanks for the help (and the reference to McKay)!
 
  • #5
Martin Rattigan said:
The sequences g1,g2,...,gp (if any) where g1g2...gp=e and it isn't the case that g1=g2=...=gp fall into sets of p that are cyclic permutations of each other, because if g1g2...gp=e then g2g3...gpg1=e etc.

If we remove these from A, since |A| = pk that leaves some multiple m=rp of p sequences ggg...g of elements such that gp=e. One of these is eee...e, so r is not zero. Therefore there must be rp-1>0 elements of order p.

(Incidentally these fall into subgroups of order p that intersect only in the identity and each contains p-1 elements of order p, so you can also say rp-1=0 (p-1) or r=1 (p-1), so that the number of elements of order p is actually
(h(p-1)+1)p-1=(hp+1)(p-1) for some integer h.)

By the way, I believe this is McKay's proof of Cauchy's theorem , first published in American Mathematical Monthly, Vol.66 (1959), page 119.

Found this thread in google (yes, old, i know), and I'm having trouble accepting this as a trivial statement (bold). Is there an easy way to just "see" this? I believe its truth rests in the fact p is prime, but it wasn't mentioned here...
 

1. What is Cauchy's Theorem and why is it important?

Cauchy's Theorem is a fundamental result in complex analysis that states that if a function is analytic on a simply connected region, then its integral along any closed contour in that region is equal to 0. This theorem is important because it allows us to easily compute integrals using complex analysis techniques.

2. What is the proof of Cauchy's Theorem?

The proof of Cauchy's Theorem involves using the Cauchy-Riemann equations and the Cauchy Integral Formula to show that the integral of an analytic function along a closed contour is equal to 0. This proof relies on the fact that analytic functions have a unique antiderivative, and thus their integrals can be easily calculated.

3. What are the assumptions and limitations of Cauchy's Theorem?

The assumptions of Cauchy's Theorem are that the function must be analytic on a simply connected region, and the contour must be closed and contained within that region. The limitations of this theorem are that it only applies to analytic functions, and the region and contour must meet the specified criteria for the theorem to hold.

4. Can Cauchy's Theorem be used to evaluate any integral?

No, Cauchy's Theorem can only be used to evaluate integrals of analytic functions along closed contours in simply connected regions. It cannot be used for integrals of non-analytic functions or along contours that do not meet the specified criteria.

5. Are there any real-world applications of Cauchy's Theorem?

Yes, Cauchy's Theorem has various real-world applications in physics, engineering, and other fields. For example, it is used in the study of fluid dynamics and electromagnetism, as well as in signal processing and control theory.

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