Prove or disprove: $S_{20}$ has an element of order 80.

Question

To be frank I don't even have a clue where to start.

I know $80$ divides $20!$, so assuming with contradiction that exists such a $g \in S_{20}$ wouldn't derive a contradiction looking at $\langle g\rangle$.

I know Cauchy's theorem states its true if 80 was prime, but since it isn't I'm not sure how to continue.

Answer 1

Sketchy answer: Find the prime decomposition of $80$, so $80=5\times 2^4$. As $5+16=21>20$, we see that $S_{20}$ contains no elements of order $80$ (but $S_{21}$ does).