element of maximal and minimal order

Question

I have to find an element of maximal order in $S_{11}$ and then find a minimal $m$ such that $α^m = 1$ for all $α ∈ S_7$. How am i supposed to solve this?

Do i have to write all the cases of $11$-cycle first?

Thank you

Answer 1

The starting point for both questions is that if $\sigma = c_1 \cdots c_k \in S_n$ is a product of $k$ disjoint cycles, then the order of $\sigma$ is $lcm(|c_i|),$ and since the cycles are disjoint, $\sum |c_i| \leq n.$

Now let $n = 11$. The question reduced to find integers $l_1, \ldots, l_k$ (representing the length of the corresponding cycles) such that $\sum l_i \leq 11$ and that $lcm(l_i)$ is maximal. Since we want to achieve the maximum, we can focus on when all $l_i \neq 1$ and and no $l_i$ is a multiple of $l_j$, since $1$'s and divisors do not increase the lcm. This leaves out relatively few options: $(9, 2), (8, 3), (7, 4), (7, 3), (7, 2), (6, 5), (6, 4), (5, 4), (5, 3), (5, 2), (5, 3, 2), (4, 3), (3, 2).$ Checking all products, you can see that $(6, 5)$ achieves the maximum of $30$.

Now let $n = 7$. We want to find $m$ such that $\alpha^m = 1$ for all $\alpha \in S_7$ (this is called the exponent of $S_7$). By definition of the order and Lagrange's theorem, if $\alpha^m = 1$, then $m$ is a multiple of the order of $\alpha$. Therefore what we are looking for is the least common multiple of all orders of elements of $S_7$. The possible disjoint cycle decompositions of elements of $S_7$ are with lengths $(7), (6), (5), (5,2), (4), (4, 3), (4, 2), (3), (3)(3), (3, 2), (3, 2, 2), (2), (2, 2), (2, 2, 2).$ These give orders $7, 6, 5, 10, 4, 12, 4, 3, 3, 6, 6, 2, 2, 2$. The least common multiple of all these numbers is $7 * 5 * 4 * 3 = 420.$