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.
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 $N=lcm(1,2,\ldots,11)=2^3\cdot 3^2\cdot 5\cdot 7\cdot 11$.