# order of element in symmetric group

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 ?

thanks

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@martin sleziak :thanks martin –  Babak Miraftab Jan 25 '12 at 16:24
This can be broken up into 2 parts: (1) Show that there exists an element of order $\prod_{i=1}^k p_k$; (2) Show that every element has order less than or equal to $\prod_{i=1}^k p_k$. Do you have a start on either of these? –  Jonas Meyer Jan 25 '12 at 16:31
@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
I agree that is the hard part. It is equivalent to showing that if $m_1+m_2+\cdots+m_j=n$ with each $m_i$ a positive integer, then $\mathrm{lcm}(m_1,m_2,\ldots,m_j)\leq \prod_{i=1}^kp_i$. –  Jonas Meyer Jan 25 '12 at 20:02
yes but i cant show this –  Babak Miraftab Jan 25 '12 at 20:10

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 $$2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23=223,092,870.$$