Wikipedia:Reference desk/Archives/Mathematics/2012 December 11

= December 11 =

Holomorph
Let H be a group of order n, K = Aut(H), and G = H x K where x is the semidirect product with respect to the identity homomorphism $$\varphi$$. Let G act on the set X of left cosets of K in G by left multiplication, inducing a permutation representation π from G into Sn. Why do we have that
 * $$\pi(G) = N_{S_n} (\pi(H))?$$

I've shown that the left side is a subset of the right, but I can't get the reverse inclusion. The textbook hints that
 * $$C_K (H) = \text{ker} \varphi = \{ 1 \}$$
 * $$C_H (K) = N_H (K)$$

could be helpful for proving
 * $$|G| = |N_{S_n} (\pi(H))|,$$

but I can't see why. Thanks in advance for any help. — Anonymous Dissident  Talk 13:54, 11 December 2012 (UTC)

Every prime has a Fibonacci divisible by it?
I'm trying to figure out the following question (The answer may be at Fibonacci number, but I'm not seeing it.) Is the following true: For all prime p there exists k such that p|Fk. If so, is there a way to calculate a k so it is true? If so, is there a way to calculate the smallest k so that it is true?


 * Yes, I think it is true. The section that you linked to says that $$F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod p$$. In other words, for prime p, one of the five Fibonacci numbers between $$F_{p-2}$$ and $$F_{p+2}$$ inclusive will be divisible by p. This is not necessarily the smallest such Fibonacci number - for example, for p = 17, $$F_{18}$$ is divisible by 17, but so is $$F_9$$. Gandalf61 (talk) 16:21, 11 December 2012 (UTC)


 * Slight correction (I think): since $$\left(\frac{p}{5}\right)$$ = 0, 1, or –1, one of the three Fibonacci numbers between $$F_{p-1}$$ and $$F_{p+1}$$ inclusive will be divisible by p. Duoduoduo (talk) 17:16, 11 December 2012 (UTC)


 * Yes. 2 divides F3 and 5 divides F5. If p is a prime of form 10n+1 or 10n+9 (meaning the last decimal digit is 1 or 9), then p divides Fp−1. If p is a prime of form 10n+3 or 10n+7 (last digit 3 or 7), then p divides Fp+1. PrimeHunter (talk) 03:39, 12 December 2012 (UTC)
 * OK, that makes sense (maybe we can put that in the article to make it more clear). This leaves my last question, is there any way to calculate the smallest Fibonacci number that p divides into? The rule says that 29 divides F28, but there is no reason one way or another that there is no n smaller than 28 for which 29 divides Fn...
 * You may also be interested in Sophie Germain primes. ~ AH1 (discuss!) 22:28, 13 December 2012 (UTC)

Reciprocals of primes
How did Euler prove the sum of the reciprocals of the primes diverges? I know he created the Riemann zeta function for it but how did he do this? — Preceding unsigned comment added by 86.185.99.78 (talk) 21:18, 11 December 2012 (UTC) — Preceding unsigned comment added by 86.185.99.78 (talk)
 * See Divergence of the sum of the reciprocals of the primes. Duoduoduo (talk) 21:35, 11 December 2012 (UTC)

These proofs are interesting but none address how he developed the Riemann zeta function to help his argument. Some of them use the sum of 1/n but none use 1/(n^s) — Preceding unsigned comment added by 86.185.99.78 (talk) 23:19, 11 December 2012 (UTC)


 * I believe he got interested in it because of the Basel problem. Dmcq (talk) 01:54, 12 December 2012 (UTC)