How to prove there is no order $15$ permutation in $S_7$?
I know that there is no element of $S_7$ that has a product of disjoint cycle with order $3$ and order $5$. However, I don't know how to write a proof of this question.
Could anyone help me? Thanks so much.
On
The cycle structure of a permutation of $S_n$ determines a partition of the integer $n$ and the order of the permutation is the $\operatorname{lcm}$ of the summands. A necessary condition for this latter to be $15$ is that $15$, or $3$ and $5$ are among the summands: the first case can happen only in $S_{n\ge 15}$, the second only in $S_{n\ge 8}$. Thence, never in $S_7$.
The order of an element of $S_7$ is the least common multiple of the cycle lengths of its cycles. For an element to have order $15$, one of its cycle lengths would have to be divisible by $3$, and one of its cycle lengths would have to be divisible by $5$. The cycle lengths divisible by $3$ are greater or equal than $3$ and the cycle lengths divisible by $5$ are greater or equal than $5$. Thus, if these are two different cycles, the sum of their lengths would have to be at least $8$, which is impossible. If they were the same cycle, its length would be divisible by $3$ and by $5$, and thus by $15$, which is also impossible.