Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $n=p_1+p_2+\cdots+p_k$ ($p_k$ is kth prime number) then $\prod_{i=1}^k p_i$ is maximum order in $S_n$.

I think it is easy but I am trying to prove it , but I have not any idea how to deal with it. any suggestions ?


share|cite|improve this question
@martin sleziak :thanks martin – Babak Miraftab Jan 25 '12 at 16:24
@jonas meyer the part 1 is not hard I can construct element with order \prod_{i=1}^k p_i but i can't show the part 2 – Babak Miraftab Jan 25 '12 at 19:54
yes but i cant show this – Babak Miraftab Jan 25 '12 at 20:10
up vote 9 down vote accepted

It isn't true in general. I found a reference to a counterexample with Google's help (specifically mentioned in this article, page 9 of the pdf file, page 359 of the publication). Namely, when $k=9$, so that$$n=100=2+3+5+7+11+13+17+19+23,$$ note that $$2^4+3^2+5+7+11+13+17+19=97<100,$$ so $S_{100}$ has an element of order $$2^4\cdot3^2\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19=232,792,560,$$ which is greater than $$2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23=223,092,870.$$

share|cite|improve this answer
(I am not sure if the link is meant to persist, so I am posting a reference to the said article.) David Gomez-Ullate and Matteo Sommacal, Periods of the Goldfish Many-Body Problem (Journal of Nonlinear Mathematical Physics). – Srivatsan Jan 25 '12 at 20:36
Thanks Srivatsan! – Jonas Meyer Jan 25 '12 at 20:38
thanks it is very helpful – Babak Miraftab Jan 26 '12 at 6:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.