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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?

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  • $\begingroup$ The lemma is just not right. One thing that would be true is that there exists $h\in G$ such that $g^h\in P$. Are you sure you have all the hypotheses of the Lemma down? $\endgroup$ – Arturo Magidin Apr 19 at 19:21
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    $\begingroup$ The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements. $\endgroup$ – Captain Lama Apr 19 at 19:21
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    $\begingroup$ Why do you think your Sylow p-subgroup $P$ is unique? $\endgroup$ – Anton Zagrivin Apr 19 at 19:25
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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.

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