I think I have a misunderstanding a part of Sylow theory for groups or I have made a big mistake in my reasoning below.
We have the following lemma in Sylow theory:
Let $G$ be a finite group and let $P$ be a Sylow-$p$ subgroup of $G$. If $g \in G$ such that the order of $g$ is $p^k$ for some $k$, then $g \in P$.
Now consider $S_5$, it has order $120 = 2^3 \times 3 \times 5$. Take a Sylow-$2$ subgroup $H$ of $S_5$. Sylow theory tells us that $|H| = 2^3 = 8$. Now any $4$-cycle in $S_5$ has order $4 = 2^2$. So any $4$-cycle in $S_5$ must be contained in $H$. But the number of $4$-cycles in $S_5$ is $\frac{5!}{(5-4)!4} = \frac{120}{4} = 30$, so $S_5$ has at least $30$ elements of order $4$ all of which must be contained in $H$ which has order $8$, an obvious contradiction.
What have I done wrong here?
The lemma you stated is false. It should be like this: if $g\in G$ has order $p^k$ for some $k\in\mathbb{N}$ then there exists a $p$-Sylow subgroup $P\leq G$ such that $g\in P$. So it doesn't say that such an element $g$ must be in all $p$-Sylow subgroups of $G$. The version that you wrote is true if there is only one $p$-Sylow subgroup.