# Misunderstanding of Sylow theory

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 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? – Arturo Magidin Apr 19 at 19:21
• The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements. – Captain Lama Apr 19 at 19:21
• Why do you think your Sylow p-subgroup $P$ is unique? – Anton Zagrivin Apr 19 at 19:25

## 1 Answer

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.