# Find the smallest integer $N$ such that all elements of $S_{11}$ have order dividing $N$.

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.

• Elements of the symmetric group can be written as a disjoint product of cycles. Can you leverage this to determine the possible orders? Jun 24 '20 at 5:02
• What is the order of $(12)(3456789)?$ Jun 24 '20 at 5:16
• @RossMillikan $l.c.m(2,7)=14$ 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 $$N=lcm(1,2,\ldots,11)=2^3\cdot 3^2\cdot 5\cdot 7\cdot 11$$.