Let $S_{11}$ be the symmetric group in 11 letters. Find (with a proof) the smallest integer $N$ such that all elements of $S_{11}$ have order dividing $N$.

I tried to find this $N$ on small order, like $S_3$, $S_4$, and $S_{5}$. For example: In $S_{4}$:

The cycles are: $I=1$

$(12)=2, (123)=3,(1234)=4, (12)(34)=2(order).$

So, I observed the minimum $N=\operatorname{lcm}(1,2,3,4)=12$.

The $S_{11}$ is a very big order group. Can anyone suggest me how I direction of the proof.

Many thanks in advance for the help.

  • 2
    $\begingroup$ Elements of the symmetric group can be written as a disjoint product of cycles. Can you leverage this to determine the possible orders? $\endgroup$ Jun 24 '20 at 5:02
  • 1
    $\begingroup$ What is the order of $(12)(3456789)?$ $\endgroup$ Jun 24 '20 at 5:16
  • $\begingroup$ @RossMillikan $l.c.m(2,7)=14$ $\endgroup$
    – User124356
    Jun 24 '20 at 5:44

As you observed, $N$ is the least common multiple of the orders of elements of $S_{11}$. Since the cycle $(1\ 2\ \cdots\ n)$ has order $n$ for $n\in\{1,2,\ldots,11\}$, we know $N$ is divisible by $lcm(1,2,\ldots,11)$. Conversely, every permutation $\sigma\in S_{11}$ is a product of disjoint cycles whose lengths are at most $11$, and the order of $\sigma$ is the least common multiple of the lengths of those cycles. Thus, the order of $\sigma$ divides $lcm(1,2,\ldots,11)$. Since $lcm(1,2,\ldots,11)$ is divisible by the order of every element of $S_{11}$ and $N$ is the least common multiple of the orders of elements of $S_{11}$, we see that $N$ divides $lcm(1,2,\ldots,11)$ and hence $N=lcm(1,2,\ldots,11)=2^3\cdot 3^2\cdot 5\cdot 7\cdot 11$.


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