Let \(S_7\) be the symmetric group on seven letters. Find all positive integers \(n\) such that some element of \(S_7\) has order \(n\).
Solution
1. If a permutation has disjoint nontrivial cycle lengths\[ d_1,\ldots,d_r, \]then its order is\[ \operatorname{lcm}(d_1,\ldots,d_r). \]
Disjoint cycles commute. The \(m\)th power of their product is the identity exactly when every cycle raised to the \(m\)th power is the identity, equivalently exactly when every \(d_i\) divides \(m\). The least positive such \(m\) is their least common multiple.
2. The possible cycle types in \(S_7\), omitting fixed points fromthe notation, give the following orders:\[ \begin{array}{c|c} \text{cycle lengths}&\text{order}\\ \hline 7&7\\ 6&6\\ 5+2&10\\ 5&5\\ 4+3&12\\ 4+2&4\\ 4&4\\ 3+3&3\\ 3+2+2&6\\ 3+2&6\\ 3&3\\ 2+2+2&2\\ 2+2&2\\ 2&2\\ \text{identity}&1 \end{array} \]
The lengths of the disjoint cycles of a permutation form a partition of \(7\) after fixed points of length \(1\) are included. Conversely every partition of \(7\) occurs as a cycle type in \(S_7\).
The displayed rows are exactly the partitions of \(7\) after suppressing the parts equal to \(1\): \[ \begin{aligned} 7,\quad 6+1,\quad 5+2,\quad 5+1+1,\quad 4+3,\quad 4+2+1,\quad 4+1+1+1,\\ 3+3+1,\quad 3+2+2,\quad 3+2+1+1,\quad 3+1+1+1+1,\\ 2+2+2+1,\quad 2+2+1+1+1,\quad 2+1+1+1+1+1,\quad 1+\cdots+1. \end{aligned} \] Applying step 1 to each row gives the stated orders.
3. The distinct positive integers occurring as element orders are\[ \boxed{ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 10,\ 12. } \]
Step 2 exhausts all possible cycle types, so no other order can occur. Each displayed number is realized by one of the listed types: respectively the identity and the types \[ 2,\quad 3,\quad 4,\quad 5,\quad 6,\quad 7,\quad 5+2,\quad 4+3. \] Thus every number in the boxed list actually occurs.
4. Q.E.D.
Step 3 gives the complete list.