Find the smallest $n$ such that $S_n$ has an element of given order.

Question

Hint given is to split $|g|$ in cycles of prime lengths.

Say, $n=1000.$

So, factored $1000$ as $2*2*2*5*5*5$ in order to find minimum value of LCM. In fact, there is no other prime factorization possible.

Say, a set of disjoint cycles given by $$(1,2)(3,4)(5,6)(7, 8, 9 ,10, 11)(12,13, 14 ,15, 16)(17, 18, 19, 20, 21)$$. But, $21$ is not correct!

Have taken a wrong approach then.

Also, why need prime length cycles in group $S_n$ to generate smallest sum as $=n.$

Not clear if this is some property of prime factorization, i.e. to generate smallest sum.

Say, if had split $1000$ into $2.4.25.5$, then would have got sum of cycle lengths as:$2+4+25+5=36.$

But, then is there a proof of this property.

Answer 1

The hint as written is somewhat misleading: Notice that all of the cycles in your permutation in $S_{21}$ have order $2$ or $5$, so the element has order $\operatorname{lcm}(2, 5) = 10$.

Since $1\,000=2^3 5^3$, we can see that to avoid this problem we'll need to find nonoverlapping cycles of lengths $2^3$ and $5^3$.

Answer 2

Hint

$\lvert g\rvert =\rm {lcm}(|c_1\rvert, \lvert c_2\rvert, \dots, \lvert c_k\rvert) $, where $g=c_1c_2\dots c_k$ is the cycle decomposition of $g$.

Write $$m=\prod_ip_i^{a_i}$$, for the prime factorization of $m$. For the purposes of this problem you can assume that $\lvert c_i\rvert =p_i^{a_i}$, (since this will give the shortest permutation with the required order).

Then we get $$n=\sum_ip_i^{a_i}$$.

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To do a different example, let $m=575$. Then $m=5^2\cdot 23$. Then $n=5^2+23=48$, so $S_{48}$ is the smallest symmetric group such that it has an element of order $575$.