Elements of order $10$ in $S_6$ 
Find the number of elements of order 10 in $S_6$

I conjecture that there are no elements of order 10 in $S_{6}$. 
If there were a $\sigma \in S_6$ s.t. $|\sigma| = 10$, then $\sigma$ is a 10 cycle, or $\sigma$ is the product of two disjoint 2 and 5-cycle. Since there is no 10-cycle or 5-cycle in $S_6$, then $\sigma$ can't be in $S_6$  
 A: The order of a permutation is the LCM of its cycle lengths. Therefore we are interested in all possible partitions of $6$, and the corresponding LCMs:
$$
\begin{align*}
6 && 6 \\
5+1 && 5 \\
4+2 && 4 \\
4+1+1 && 4 \\
3+3 && 3 \\
3+2+1 && 6 \\
3+1+1+1 && 3 \\
2+2+2 && 2 \\
2+2+1+1 && 2 \\
2+1+1+1+1 && 2 \\
1+1+1+1+1+1 && 1
\end{align*}
$$
Therefore the possible orders are $1,2,3,4,5,6$.
There is no real need to look at the entire list in this case. If the LCM of the cycle sizes is 10, then one of the cycle lengths needs to be a multiple of 5. The only multiply of 5 which is at most 6 is 5, and the only corresponding partition is 5+1, with LCM 5.
Similarly, let us show that in $S_{17}$ there is no permutation of order 65. Since $65 = 13\times 5$, we would need to have a cycle of length 13, which leaves no room for a cycle whose length is a multiple of 5.
A: Let $\sigma$ be a permutation of order $10$ , $\sigma$ has at least one cycle of even length and one cycle of length multiple of $5$ (It can be the same cycle but then we need at least $10$ elements in that cycle). If it isn't the same then the even has size at least $2$ and the multiple of $5$ at least $5$. Hence $S_7$ is the smallest symmetric group with elements of order $10$.
