I am having trouble with the following problem:
Show that $S_n$ has elements of order $p^t \Longleftrightarrow n \geq p ^t$, being $n, t$ positive integers and $p$ a prime number.
Here is what I thought:
Let $\sigma \in S_n$ have order $p^t$. We can write $\sigma$ as a product of disjoint cycles of lenght $>1$: $$ \sigma = \sigma_1 \dots \sigma_r $$ Then $p^t = $ lcm $(o(\sigma_1), \dots, o(\sigma_r))$, being $o(\sigma_i)$ the order of $\sigma_i$. Since $p$ is a prime and we are only considering nontrivial cycles, we must have $o(\sigma_i) = p^{n_i}$. Now, since the maximum length in $S_n$ is $n$, we must have $$ p^{n_1} + \dots + p^{n_r} \leq n $$ And here I am stuck. How can I relate $p^t$ to the above sum? Or is there another way?
Thanks in advance.