# Prove that a permutation of size N can or cannot be of order M

An example of this would be a permutation of size 32. Can this be of order 62? I assume not because the only way to achieve this would be to have the permutation be made up out of a part with order 2 and one with order 31, but this is impossible.

Am I correct, and what is a formal way of proving this?

Therefore, a permutation of size $$n$$ can be of order $$m$$ iff there exist numbers $$p_1,p_2,\dots,p_k$$ which sum to at most $$n$$ and multiply to $$m$$. Clearly, this remains true if the $$p_i$$ are required to be distinct coprime numbers greater than $$1$$, which can be used to simplify proofs of (non-)existence.
Each permutation of size $$n\geq 1$$ can be written as the product of $$k\geq 1$$ disjoint cycles $$C_1,C_2,\dots,C_k$$ and the order of $$P$$ is the least common multiple of the lengths of the cycles.
Hence, there is a permutation of size $$n$$ of order $$m$$ if and only $$\begin{cases} l_1+l_2+\dots+l_k=n\\ \text{lcm}(l_1,l_2,\dots,l_k)=m \end{cases}$$ has a solution with $$1\leq l_1\leq l_2\leq \dots\leq l_k$$ positive integers and $$k\geq 1$$.
In your case $$n=32$$ and $$m=62$$. Since $$62=2\cdot 31$$ and $$1\leq l_i\leq 32$$, it follows easily that there are no solutions. So, yes, you are correct.